Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 17 Circles

Exercise 17A

Question 1. In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. Find ∠AOC. Show your steps of working.
Answer:

[Diagram: This diagram shows a circle with center O and three points A, B, C on the circumference. Lines OA, OB, OC are drawn from center to points. Angle OAB is marked as 30° and angle OCB is marked as 40°.]

Step 1: Join AC.
Let ∠OAC = ∠OCA = x (say)

Step 2: Find ∠AOC using triangle OAC.
∠AOC = 180° - 2x

Step 3: Find angles in triangles.
Also, ∠BAC = 30° + x
∠BCA = 40° + x

Step 4: Use angle sum property in triangle ABC.
In △ABC,
∠ABC = 180° - ∠BAC - ∠BCA
= 180° - (30° + x) - (40° + x) = 110° - 2x

Step 5: Apply the theorem that angle at center is double the angle at circumference.
Now, ∠AOC = 2∠ABC
180° - 2x = 2(110° - 2x)
180° - 2x = 220° - 4x
2x = 40°
x = 20°

Step 6: Calculate final answer.
∠AOC = 180° - 2 × 20° = 140°

In simple words: We used the fact that the angle at the center is twice the angle at the edge of the circle. We found the unknown angle by solving step by step.

📝 Teacher's Note: Draw the extra line AC to make two triangles. This helps students see the relationship between center angle and circumference angle clearly.

🎯 Exam Tip: Always write "angle at center = 2 × angle at circumference" clearly. Show all steps. Don't skip the triangle properties.

Question 2. In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
(i) Prove that AC is a diameter of the circle.
(ii) Find ∠ACB.
Answer:

[Diagram: This diagram shows a circle with four points A, B, C, D on the circumference. Various angles are marked including BAD = 65°, ABD = 70°, and BDC = 45°.]

(i) To prove AC is a diameter:
In △ABD,
∠DAB + ∠ABD + ∠ADB = 180°
⇒65° + 70° + ∠ADB = 180°
⇒135° + ∠ADB = 180°
⇒ ∠ADB = 180° - 135° = 45°
Now, ∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°
Since ∠ADC is the angle of semicircle, so AC is a diameter of the circle.

(ii) Finding ∠ACB:
∠ACB = ∠ADB ....(angles in the same segment of a circle)
⇒ ∠ACB = 45°

In simple words: When the angle in a semicircle is 90°, the line across is a diameter. We used angle properties of triangles and circles to solve this.

📝 Teacher's Note: Remind students that angle in semicircle is always 90°. If they find 90° angle, the line across must be diameter. This is a key circle theorem.

🎯 Exam Tip: Write "angle in semicircle = 90°, therefore AC is diameter" clearly. Also write "angles in same segment are equal" for part (ii).

Question 3. Given O is the centre of the circle and ∠AOB = 70°. Calculate the value of:
(i) ∠OCA,
(ii) ∠OAC.
Answer:

[Diagram: This diagram shows a circle with center O and three points A, B, C on the circumference. The central angle AOB is marked as 70°.]

Here, ∠AOB = 2∠ACB
(Angle at the centre is double the angle at the circumference subtended by the same chord)
⇒ ∠ACB = \(\frac{70°}{2}\) = 35°
Now, OC = OA (Radii of same circle)
⇒ ∠OCA = ∠OAC = 35°

In simple words: Since OA and OC are both radii, triangle OAC has two equal sides. So the two base angles are equal. We found this angle using the center-circumference angle rule.

📝 Teacher's Note: Show students that radii are always equal. This makes isosceles triangles with equal base angles. Draw this clearly on the board.

🎯 Exam Tip: Write "OA = OC (radii)" and "angle at center = 2 × angle at circumference". These key statements get you marks.

Question 4. In each of the following figures, O is the centre of the circle. Find the values of a, b, and c.
Answer:

[Diagram: Two circle diagrams labeled (i) and (ii). In (i), there's a central angle of 130° and angles marked as 'a' and 'b'. In (ii), there's a central angle of 112° and an angle marked as 'c'.]

(i) Here, b = \(\frac{1}{2}\) × 130°
(Angle at the centre is double the angle at the circumference subtended by the same chord)
⇒ b = 65°
Now, a + b = 180°
(Opposite angles of a cyclic quadrilateral are supplementary)
⇒ a = 180° - 65° = 115°

(ii) Here, c = \(\frac{1}{2}\) Reflex{112°}
(Angle at the centre is double the angle at the circumference subtended by the same chord)
⇒ c = \(\frac{1}{2}\) × (360° - 112°) = 124°

In simple words: We used two main rules - center angle is double the edge angle, and opposite angles in a cyclic shape add to 180°. For reflex angles, we subtract from 360° first.

📝 Teacher's Note: Explain that reflex angle means the bigger angle (more than 180°). Show students how to find it by subtracting the smaller angle from 360°.

🎯 Exam Tip: For cyclic quadrilaterals, write "opposite angles are supplementary" clearly. For reflex angles, show the calculation 360° - given angle.

Question 5. In each of the following figures, O is the centre of the circle. Find the value of a, b, c and d.
Answer:

[Diagram: Four circle diagrams labeled (i), (ii), (iii), and (iv) showing various angles and central points with different angle measurements marked.]

(i)
∠AOB = 70° + 35° = 105° (angles on a straight line)
a = \(\frac{1}{2}\) × 105° = 52.5°
(Angle at center is double the angle at circumference)

(ii)
∠AOB = 120° (given at center)
b = \(\frac{1}{2}\) × 120° = 60°
c = 25° (inscribed angle given)
∠ADB = 180° - 120° = 60° (angle in semicircle concept)

(iii)
∠ACB = 50° (given at center)
c = \(\frac{1}{2}\) × 50° = 25°
(Angle at circumference is half the center angle)

(iv)
∠APB = 45° (given)
∠AOB = 2 × 45° = 90° (center angle is double)
d = \(\frac{1}{2}\) × 90° = 45°

In simple words: We used the main circle rule that center angle equals twice the edge angle. We also used the fact that angles in the same segment are equal.

📝 Teacher's Note: Practice these problems step by step. Students often forget to identify which is the center angle and which is the circumference angle. Point this out clearly.

🎯 Exam Tip: Always identify center angles and circumference angles first. Write the 2:1 relationship clearly. Show your working for each part separately.

[Diagram: Two circles showing triangles inscribed in them - figure (iii) shows triangle ABC in a circle with center O and angle measurements, figure (iv) shows triangle ABC in a circle with center O and angle measurements]

(i) Here, ∠BAD = 90° (Angle in a semicircle)
Answer: ∠BDA = 90° - 35° = 55°
Again, a = ∠ACB = ∠BDA = 55°
(Angles subtended by the same chord on the circle are equal)
In simple words: When an angle is made in a semicircle, it is always 90°. Then we use triangle angle sum to find other angles.

📝 Teacher's Note: Show students that any angle in a semicircle is always 90°. This is a very important circle rule. Draw different semicircles on the board to make this clear.

🎯 Exam Tip: Always write "angle in a semicircle = 90°" first. Then use triangle angle sum. Examiners look for this key phrase.

(ii) Here, ∠DAC = ∠CBD = 25° (Angles subtended by the same chord on the circle are equal)
Answer: Again, 120° = b + 25°
(In a triangle, measure of exterior angle is equal to the sum of pair of opposite interior angles)
∴ b = 95°
In simple words: Equal chords make equal angles. Then we use the exterior angle rule - outside angle equals sum of two inside opposite angles.

📝 Teacher's Note: Draw a triangle and show how the exterior angle equals the sum of the two non-adjacent interior angles. This rule helps solve many circle problems.

🎯 Exam Tip: Write "angles subtended by same chord are equal" and "exterior angle = sum of opposite interior angles". These exact phrases get marks.

(iii) ∠AOB = 2∠AOB = 2 × 50° = 100° (Angle at the centre is double the angle at the circumference subtended by the same chord)
Answer: Also, OA = OB
∴ ∠OBA = ∠OAB = c
∴ c = \(\frac{180° - 100°}{2}\) = 40°
In simple words: The angle at the center is twice the angle at the edge. Since OA and OB are radii, triangle OAB has two equal sides, so two equal angles.

📝 Teacher's Note: Explain that center angle is double the circumference angle. Also show that radii are equal, so the triangle formed is isosceles with two equal angles.

🎯 Exam Tip: Write "angle at center = 2 × angle at circumference" and "radii are equal, so isosceles triangle". Both statements get marks.

(iv) ∠APB = 90° (Angle in a semicircle)
Answer: ∠BAP = 90° - 45° = 45°
Now, d = ∠BCP = ∠BAP = 45°
(Angles subtended by the same chord on the circle are equal)
In simple words: Any angle in a semicircle is 90°. Equal chords make equal angles anywhere on the circle.

📝 Teacher's Note: Remind students that semicircle angles are always 90°. Then show how the same chord makes equal angles at different points on the circle.

🎯 Exam Tip: First write "angle in semicircle = 90°", then "angles subtended by same chord are equal". This order is important for full marks.

Question 6. In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O₁ and O₂ are the centres of two circles.
Answer:

[Diagram: Two intersecting circles with centers O₁ and O₂, showing points A, B, C, D where AB is a common chord and AC, AD are diameters]

∠DBA = 90° and ∠CBA = 90°
(Angles in a semicircle is a right angle)
Adding both we get,
∠DBC = 180°
∴ D, B and C form a straight line.
In simple words: Since AC and AD are diameters, angles DBA and CBA are both 90°. When two 90° angles are next to each other, they make 180°, which means the points are in a straight line.

📝 Teacher's Note: Draw two semicircles sharing a common chord. Show that both angles at B are 90°. When these add to 180°, the three points must be in a line.

🎯 Exam Tip: Write "angle in semicircle = 90°" for both circles. Then write "90° + 90° = 180°, so points are collinear". Use the word "collinear".

Question 7. In the figure given below, find: (i) ∠BCD, (ii) ∠ADC, (iii) ∠ABC.
Answer:

[Diagram: Circle with cyclic quadrilateral ABCD where angle BAD = 105°]

(i) ∠BCD + ∠BAD = 180°
(Sum of opposite angles of a cyclic quadrilateral is 180°)
∴ ∠BCD = 180° - 105° = 75°

(ii) Now, AB || CD
∴ ∠BAD + ∠ADC = 180°
(Interior angles on same side of parallel lines is 180°)
∴ ∠ADC = 180° - 105° = 75°

(iii) ∠ADC + ∠ABC = 180°
(Sum of opposite angles of a cyclic quadrilateral is 180°)
∴ ∠ABC = 180° - 75° = 105°
In simple words: In a cyclic quadrilateral (four-sided shape in a circle), opposite angles always add up to 180°. We use this rule to find the missing angles.

📝 Teacher's Note: Draw a quadrilateral inside a circle. Show students that opposite angles (across from each other) always add to 180°. This is the key rule for cyclic quadrilaterals.

🎯 Exam Tip: Always write "opposite angles of cyclic quadrilateral = 180°" for each calculation. This phrase is essential for marks. Show the working step by step.

Question 8. In the given figure, O is centre of the circle. If ∠AOB = 140° and ∠OAC = 50°, find: (i) ∠ACB, (ii) ∠OBC, (iii) ∠OAB, (iv) ∠CBA
Answer:

[Diagram: Circle with center O, showing points A, B, C where angle AOB = 140° and angle OAC = 50°]

Here, ∠ACB = \(\frac{1}{2}\) Reflex {∠AOB} = \(\frac{1}{2}\){360° - 140°} = 110°
(Angle at the centre is double the angle at the circumference subtended by the same chord)
Now, OA = OB (Radii of same circle)
∴ ∠OBA = ∠OAB = \(\frac{180° - 140°}{2}\) = 20°
∴ ∠CAB = 50° - 20° = 30°
In ΔCAB,
∠CBA = 180° - 110° - 30° = 40°
∴ ∠OBC = ∠CBA + ∠OBA = 40° + 20° = 60°
In simple words: We use the center angle rule (center angle is double the circumference angle). Since OA and OB are equal radii, triangle OAB has two equal angles.

📝 Teacher's Note: Show that when the center angle is more than 180°, we use the reflex angle. Draw the isosceles triangle OAB to show the two equal angles at the base.

🎯 Exam Tip: When center angle is more than 180°, use reflex angle in the formula. Write "radii are equal" to show the isosceles triangle. Show all angle calculations step by step.

Question 9. Calculate: (i) ∠CDB, (ii) ∠ABC, (iii) ∠ACB.
Answer:

[Diagram: Circle with cyclic quadrilateral ABCD showing angles 43° and 49°]

Here,
∠CDB = ∠BAC = 49°
∠ABC = ∠ADC = 43°
(Angles subtended by the same chord on the circle are equal)
By angle - sum property of a triangle,
∠ACB = 180° - 49° - 43° = 88°
In simple words: Equal chords make equal angles anywhere on the circle. Then we use triangle angle sum rule (all angles in a triangle add to 180°) to find the third angle.

📝 Teacher's Note: Show students how the same chord creates equal angles at different points on the circle. Then use the basic triangle rule that all angles add to 180°.

🎯 Exam Tip: Write "angles subtended by same chord are equal" first. Then write "angle sum in triangle = 180°". Both statements are needed for full marks.

Question 10. In the figure given below, ABCD is a cyclic quadrilateral in which ∠ BAD = 75°; ∠ ABD = 58° and ∠ADC = 77°. Find:
(i) ∠ BDC,
(ii) ∠ BCD,
(iii) ∠ BCA.

[Diagram: This diagram shows a cyclic quadrilateral ABCD inscribed in a circle, with angles marked as 75°, 58°, and 77°.]

Answer:
(i) By angle-sum property of triangle ABD,
∠ADB = 180° - 75° - 58° = 47°
\( \implies \) ∠BDC = ∠ADC - ∠ADB = 77° - 47° = 30°

(ii) ∠BAD + ∠BCD = 180°
(Sum of opposite angles of a cyclic quadrilateral is 180°)
\( \implies \) 75° + ∠BCD = 180°
\( \implies \) ∠BCD = 180° - 75° = 105°

(iii) ∠BCA = ∠ADB = 47°
(Angles subtended by the same chord on the circle are equal)

In simple words: In a cyclic quadrilateral, opposite angles add up to 180°. Also, angles made by the same chord at different points on the circle are equal. We use these rules to find all the missing angles.

📝 Teacher's Note: Draw a circle on the board and mark four points. Show students that opposite angles always add to 180°. This is the most important property of cyclic quadrilaterals.

🎯 Exam Tip: Always write "Sum of opposite angles = 180°" and "Angles subtended by same chord are equal." These key phrases get you marks even if calculations are wrong.

Question 11. In the following figure, O is centre of the circle and ∆ ABC is equilateral. Find:
(i) ∠ ADB
(ii) ∠ AEB

[Diagram: This diagram shows a circle with center O, containing an equilateral triangle ABC, with additional points D and E on the circle.]

Answer:
Since ∠ACB and ∠ADB are in the same segment,
∠ADB = ∠ACB = 60°
Join OA and OB.
Here, ∠AOB = 2∠ACB = 2 × 60° = 120°
∠AEB = \( \frac{1}{2} \) Reflex(∠AOB) = \( \frac{1}{2} \)(360° - 120°) = 120°
(Angle at the centre is double the angle at the circumference subtended by the same chord)

In simple words: In an equilateral triangle, all angles are 60°. When this triangle is inside a circle, we can use circle properties to find angles at other points on the circle.

📝 Teacher's Note: First make students remember that equilateral triangle has all angles = 60°. Then teach the circle property: angle at center = 2 × angle at circumference.

🎯 Exam Tip: Write "equilateral triangle has all angles = 60°" first. Then use circle theorems. Show both steps clearly for full marks.

Question 12. Given—∠ CAB = 75° and ∠ CBA = 50°. Find the value of ∠ DAB + ∠ ABD

[Diagram: This diagram shows a circle with points A, B, C, and D marked, with angles 75° and 50° labeled.]

Answer:
In ∆ABC, ∠CBA = 50°, ∠CAB = 75°
∠ACB = 180° - (∠CBA + ∠CAB)
= 180° - (50° + 75°)
= 180° - 125°
= 55°
But ∠ADB = ∠ACB = 55°
(Angles subtended by the same chord on the circle are equal)
Now consider ∆ABD,
∠DAB + ∠ABD + ∠ADB = 180°
\( \implies \) ∠DAB + ∠ABD + 55° = 180°
\( \implies \) ∠DAB + ∠ABD = 180° - 55°
\( \implies \) ∠DAB + ∠ABD = 125°

In simple words: First we find the third angle of triangle ABC. Then we use the rule that angles made by the same chord are equal. Finally we use the triangle angle sum rule.

📝 Teacher's Note: Show students step by step: first complete the triangle, then use circle properties, then use triangle properties again. This teaches them to combine different rules.

🎯 Exam Tip: Write each step clearly. First find angle ACB, then use "angles subtended by same chord", then use angle sum in triangle ABD. Show all working.

Question 13. ABCD is a cyclic quadrilateral in a circle with centre O. If ∠ ADC = 130°; find ∠ BAC.

[Diagram: This diagram shows a cyclic quadrilateral ABCD in a circle with center O marked.]

Answer:
Here, ∠ACB = 90°
(Angle in a semicircle is a right angle)
Also, ∠ABC = 180° - ∠ADC = 180° - 130° = 50°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
By angle sum property of right triangle ACB,
∠BAC = 90° - ∠ABC = 90° - 50° = 40°

In simple words: In a cyclic quadrilateral, opposite angles add up to 180°. We use this to find one angle, then use triangle properties to find the required angle.

📝 Teacher's Note: Remind students that angle in semicircle is always 90°. Also teach that opposite angles in cyclic quadrilateral are supplementary (add to 180°).

🎯 Exam Tip: Write "opposite angles are supplementary" and "angle in semicircle = 90°". These are the key properties that solve this question. Show the calculation clearly.

Question 14. In the figure given below, AOB is a diameter of the circle and ∠ AOC = 110°. Find ∠ BDC.

[Diagram: This diagram shows a circle with diameter AOB, point C on the circle, angle AOC marked as 110°, and point D on the circle.]

Answer:
Join AD.
Here, ∠ADC = \( \frac{1}{2} \)∠AOC = \( \frac{1}{2} \) × 110° = 55°
(Angle at the centre is double the angle at the circumference subtended by the same chord)
Also, ∠ADB = 90°
(Angle in a semicircle is a right angle)
\( \implies \) ∠BDC = 90° - ∠ADC = 90° - 55° = 35°

In simple words: We use two circle rules here. First, angle at center is double the angle at circumference. Second, any angle in a semicircle is 90°. We combine these to find the answer.

📝 Teacher's Note: Show students that AOB is a diameter, so any angle ADB must be 90°. Then show how center angle and circumference angle are related. Draw these clearly on board.

🎯 Exam Tip: Write "angle in semicircle = 90°" and "angle at center = 2 × angle at circumference". Use both properties. Show that you joined AD to create the semicircle.

Question 15. In the following figure, O is centre of the circle, ∠AOB = 60° and ∠BDC = 100°. Find ∠OBC.
Answer:
Given:
O is centre of the circle
∠AOB = 60°
∠BDC = 100°

Step 1: Find ∠ACB using angle at centre theorem.
∠ACB = \( \frac{1}{2} \) × ∠AOB = \( \frac{1}{2} \) × 60° = 30°
(Angle at the centre is double the angle at the circumference subtended by the same chord)

Step 2: Use angle sum property of quadrilateral ABDC.
In quadrilateral ABDC: ∠DBC = 180° - 100° - 30° = 50°

Therefore, ∠OBC = 50°
In simple words: We used the rule that the centre angle is double the edge angle. Then we used the fact that opposite angles in a four-sided shape add up to 180°.

[Diagram: Circle with centre O, points A, B, C, D on circumference, showing angles 60° at centre and 100° at circumference]

📝 Teacher's Note: Draw a pizza slice to show centre angle. Then show how edge angle is half of centre angle. Students remember this pizza slice rule easily.

🎯 Exam Tip: Always write "angle at centre is double the angle at circumference" for full marks. Show all working steps clearly.

Question 16. ABCD is a cyclic quadrilateral in which ∠DAC = 27°; ∠DBA = 50° and ∠ADB = 33°. Calculate: (i) ∠DBC, (ii) ∠DCB, (iii) ∠CAB.
Answer:
(i) ∠DBC = ∠DAC = 27°
(Angles subtended by the same chord on the circle are equal)

(ii) ∠ACB = ∠ADB = 33°
(Angles subtended by the same chord on the circle are equal)
∠ACD = ∠ABD = 50°
(Angles subtended by the same chord on the circle are equal)
∴ ∠DCB = ∠ACD + ∠ACB = 50° + 33° = 83°

(iii) In triangle ADB:
∠DAB + ∠DBA + ∠ADB = 180°
(Sum of angles in triangle = 180°)
⇒ ∠DAB = 180° - 50° - 33° = 97°
But ∠DAB = ∠DAC + ∠CAB = 27° + ∠CAB
⇒ 97° = 27° + ∠CAB
⇒ ∠CAB = 97° - 27° = 70°

Final Answers:
(i) ∠DBC = 27°
(ii) ∠DCB = 83°
(iii) ∠CAB = 70°
In simple words: In a circle, angles made by the same curved part are equal. We used this rule and triangle angle sum to find all angles.

📝 Teacher's Note: Show students a circle with a chord. Mark different points on the circle. All angles made by that chord are equal - this is the key rule.

🎯 Exam Tip: Write "angles subtended by same chord are equal" for each step. Also remember triangle angle sum is always 180°.

Question 17. In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°. Find the number of degrees in: (i) ∠DCE; (ii) ∠ABC.
Answer:
(i) ∠CED = 90°
(Angle in a semicircle is a right angle)
∴ ∠DCE = 90° - ∠CDE = 90° - 40° = 50°
∴ ∠DCE = ∠OCB = 50°

(ii) In triangle BOC:
∠AOC = ∠OCB + ∠OBC
(Exterior angle of triangle is equal to the sum of pair of interior opposite angles)
⇒ ∠OBC = 80° - 50° = 30° [∠AOC = 80°, given]

Hence, ∠ABC = 30°
In simple words: Any angle in a half circle is always 90°. We used this rule and the fact that outside angle equals sum of two inside opposite angles.

[Diagram: Circle with centre O, diameters AB and CD, point E on circumference, showing angles 80° and 40°]

📝 Teacher's Note: Cut a semicircle from paper. Show students that any triangle in a semicircle always has one 90° angle. This is a very important rule.

🎯 Exam Tip: Write "angle in semicircle = 90°" clearly. Also mention "exterior angle theorem" when you use it. These key phrases get you marks.

Question 17 (old). In the figure given below, AB is diameter of the circle whose centre is O. Given that: ∠ECD = ∠EDC = 32°. Show that ∠COF = ∠CEF.
Answer:
Step 1: Find ∠COF using centre angle theorem.
∠COF = 2∠CDF = 2 × 32° = 64° --- (i)
(Angle at the centre is double the angle at the circumference subtended by the same chord)

Step 2: Find ∠CEF in triangle ECD.
In triangle ECD:
∠CEF = ∠ECD + ∠EDC = 32° + 32° = 64° --- (ii)
(Exterior angle of triangle is equal to the sum of pair of interior opposite angles)

Step 3: Compare equations (i) and (ii).
From (i) and (ii), we get
∠COF = ∠CEF

Hence proved.
In simple words: We found both angles using different circle rules. The centre angle rule and the outside angle rule gave us the same answer, so they are equal.

[Diagram: Circle with diameter AB, centre O, points C, D, E, F on circumference, showing equal angles of 32°]

📝 Teacher's Note: Explain that "prove" means show both sides are equal using circle rules. Use two different methods to find the same angle.

🎯 Exam Tip: Number your steps clearly. Write the circle theorem you used in brackets. End with "Hence proved" for proof questions.

Question 18. In the figure given below, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that AB = BE.
Answer:
Step 1: Join OB.
Then ∠OBA = 90°
(Angle in a semicircle is a right angle)
i.e. OB ⊥ AE

Step 2: Apply perpendicular from centre theorem.
We know, the perpendicular drawn from the centre to a chord bisects the chord.
∴ AB = BE

Hence proved.
In simple words: We joined the centre O to point B. This made a 90° angle with line AE. When a line from centre is perpendicular to a chord, it cuts the chord into two equal parts.

[Diagram: Two circles - larger circle with centre O and diameter AC, smaller circle with diameter AO, chord AE intersecting smaller circle at B]

📝 Teacher's Note: Draw a circle and any chord. Show that the perpendicular from centre always cuts the chord exactly in half. This is a very useful rule.

🎯 Exam Tip: Write "perpendicular from centre bisects chord" as your key theorem. Show that OB ⊥ AE clearly in your working.

Question 19. In the following figure, (i) if ∠BAD = 96°, find BCD and (ii) Prove that AD is parallel to FE.
Answer:
(i) Finding ∠BCD:
In cyclic quadrilateral ABCD:
∠BAD + ∠BCD = 180°
(Opposite angles of cyclic quadrilateral are supplementary)
96° + ∠BCD = 180°
∠BCD = 180° - 96° = 84°

(ii) Proving AD || FE:
∠BAD = 96° (given)
∠BFE = ∠BAD = 96°
(Angles subtended by the same arc are equal)
Since ∠BAD = ∠BFE and they are corresponding angles,
AD || FE

Hence proved.
In simple words: In a four-sided shape inside a circle, opposite angles add up to 180°. When two angles are equal and in the right position, the lines are parallel.

[Diagram: Two intersecting circles with points A, B, C, D, E, F marked, showing angle of 96°]

📝 Teacher's Note: Show students a cycle quadrilateral drawn in a box. Opposite corners always add to 180°. For parallel lines, equal corresponding angles is the key rule.

🎯 Exam Tip: For cyclic quadrilateral, write "opposite angles are supplementary". For parallel lines, write "corresponding angles are equal". Use these exact phrases.

Question 20. Prove that:
(i) the parallelogram, inscribed in a circle, is a rectangle.
(ii) the rhombus, inscribed in a circle, is a square.

Solution:

[Diagram: Shows a parallelogram ABCD inscribed in a circle, with angles marked including 96°]

(i) Let ABCD be a parallelogram, inscribed in a circle.
Now, \( \angle BAD = \angle BCD \)
(Opposite angles of a parallelogram are equal)
and \( \angle BAD + \angle BCD = 180° \)
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
\( \therefore \angle BAD = \angle BCD = \frac{180°}{2} = 90° \)
Similarly, the other two angles are \( 90° \) and opposite pair of sides are equal.
\( \therefore \) ABCD is a rectangle.

(ii) Let ABCD be a rhombus, inscribed in a circle.
Now, \( \angle BAD = \angle BCD \)
(opposite angles of a rhombus are equal)
and \( \angle BAD + \angle BCD = 180° \)
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
\( \therefore \angle BAD = \angle BCD = \frac{180°}{2} = 90° \)
Similarly, the other two angles are \( 90° \) and all the sides are equal.
\( \therefore \) ABCD is a square.

In simple words: When a parallelogram is drawn inside a circle, all its angles become 90°. This makes it a rectangle. When a rhombus is drawn inside a circle, all angles become 90°, making it a square.

📝 Teacher's Note: Show students that any quadrilateral inside a circle has opposite angles adding to 180°. This special property forces parallelograms to become rectangles and rhombuses to become squares.

🎯 Exam Tip: Always write the property "opposite angles in cyclic quadrilateral are supplementary" clearly. Then show how this makes all angles 90°. This gets full marks.

Question 21. In the following figure, AB = AC. Prove that DECB is an isosceles trapezium.

[Diagram: Shows triangle ABC inscribed in a circle with points D and E on the circle, forming quadrilateral DECB]

Solution:

Here, AB = AC
\( \Rightarrow \angle B = \angle C \)
\( \therefore \) DECB is a cyclic quadrilateral.
(In a triangle, angles opposite to equal sides are equal)
Also, \( \angle B + \angle DEC = 180° \) ... (1)
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
\( \Rightarrow \angle C + \angle DEC = 180° \) [from (1)]
But this is the sum of interior angles on one side of a transversal.
\( \therefore \) DE || BC
But \( \angle ADE = \angle B \) and \( \angle AED = \angle C \) [corresponding angles]
Thus, \( \angle ADE = \angle AED \)
\( \Rightarrow \) AD = AE
\( \Rightarrow \) AB - AD = AC - AE (\( \because \) AB = AC)
\( \Rightarrow \) BD = CE
Thus, we have, DE || BC and BD = CE
Hence, DECB is an isosceles trapezium.

In simple words: We proved that DE is parallel to BC (making it a trapezium) and the non-parallel sides BD and CE are equal (making it isosceles).

📝 Teacher's Note: Explain that an isosceles trapezium has one pair of parallel sides and the other pair of sides are equal. Use the properties of cyclic quadrilaterals to prove both conditions.

🎯 Exam Tip: Write clearly "DE || BC" to show it's a trapezium, then "BD = CE" to show it's isosceles. These are the two key properties examiners look for.

Question 22. Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.

[Diagram: Shows two intersecting circles with diameters PA and PB drawn through point P, and point Q at the other intersection]

Solution:

Let O and O' be the centres of two intersecting circles, where points of intersection are P and Q and PA and PB are their diameters respectively.
Join PQ, AQ and QB.
\( \therefore \angle AQP = 90° \) and \( \angle BQP = 90° \)
(Angle in a semicircle is a right angle)
Adding both these angles,
\( \angle AQP + \angle BQP = 180° \Rightarrow \angle AQB = 180° \)
Hence, the points A, Q and B are collinear.

In simple words: Since PA and PB are diameters, angles AQP and BQP are both 90°. When we add them, we get 180°, which means A, Q, and B are on a straight line.

📝 Teacher's Note: Use the property that angle in semicircle is 90°. When two right angles meet at a point, they form a straight line (180°).

🎯 Exam Tip: Always state "angle in semicircle is 90°" first. Then show that \( \angle AQP + \angle BQP = 180° \) proves collinearity. This is the complete proof.

Question 23. The figure given below, shows a circle with centre O. Given: \( \angle AOC = a \) and \( \angle ABC = b \).
(i) Find the relationship between a and b
(ii) Find the measure of angle OAB, if OABC is a parallelogram.

[Diagram: Shows a circle with centre O, with points A, B, C on the circle, showing central angle a and inscribed angle b]

Solution:

(i) \( \angle ABC = \frac{1}{2} \) Reflex \( \angle COA \)
(Angle at the centre is double the angle at the circumference subtended by the same chord)
\( \Rightarrow b = \frac{1}{2}(360° - a) \)
\( \Rightarrow a + 2b = 180° \)

(ii) Since OABC is a parallelogram, so opposite angles are equal
\( \therefore a = b \)
Using relationship in (i),
\( 3a = 180° \)
\( \therefore a = 60° \)
Also, OC || BA
\( \therefore \angle COA + \angle OAB = 180° \)
\( \Rightarrow 60° + \angle OAB = 180° \)
\( \Rightarrow \angle OAB = 120° \)

In simple words: The central angle is twice the inscribed angle. When OABC is a parallelogram, we use this relationship along with parallel line properties to find the angle.

📝 Teacher's Note: Remind students that central angle = 2 × inscribed angle. In parallelograms, opposite angles are equal and adjacent angles are supplementary.

🎯 Exam Tip: For part (i), write the central angle theorem clearly. For part (ii), use "opposite angles equal in parallelogram" and "co-interior angles supplementary" properties.

Question 24. Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD at the centre O is equal to twice the angle APC.

Solution:

[Diagram: Shows a circle with centre O, two chords AB and CD intersecting at point P inside the circle, creating arcs AC and BD]

Let the angle subtended by arc AC at centre O be \( \alpha \) and by arc BD be \( \beta \).
We need to prove: \( \alpha + \beta = 2 \angle APC \)

In triangle APC:
Exterior angle at P = Sum of two interior opposite angles
\( \angle APC = \frac{1}{2} \) (arc AC) + \frac{1}{2} \) (arc BD)
(Angle formed by two chords inside circle = \( \frac{1}{2} \) sum of intercepted arcs)
\( \angle APC = \frac{1}{2}(\alpha + \beta) \)
\( \therefore \alpha + \beta = 2 \angle APC \)

Hence proved.

In simple words: When two chords cross inside a circle, the angle they make equals half the sum of the arcs they cut off. So the sum of those arcs equals twice that angle.

📝 Teacher's Note: This is the intersecting chords angle theorem. Draw the diagram clearly and show students how the angle at P relates to the two arcs cut off by the chords.

🎯 Exam Tip: State the theorem clearly: "Angle formed by intersecting chords = \( \frac{1}{2} \) sum of intercepted arcs." Then rearrange to get the required result. This shows complete understanding.

Question 24 (old). ABCD is a quadrilateral inscribed in a circle having ∠A = 60°; O is the centre of the circle. Show that: ∠OBD + ∠ODB = ∠CBD + ∠CDB

[Diagram: This diagram shows a circle with center O. Inside the circle is a quadrilateral ABCD with all four corners touching the circle. Lines are drawn from O to all four corners, and also some diagonal lines inside the quadrilateral.]

Answer:

Given: Two chords AB and CD intersect each other at P inside the circle. OA, OB, OC and OD are joined.

To prove: \( \angle AOC + \angle BOD = 2\angle APC \)

Construction: Join AD.

Proof:

Arc AC subtends \( \angle AOC \) at the centre and \( \angle ADC \) at the remaining part of the circle.

\( \angle AOC = 2\angle ADC \) ......(1)

Similarly, \( \angle BOD = 2\angle BAD \) ......(2)

Adding (1) and (2),

\( \angle AOC + \angle BOD = 2\angle ADC + 2\angle BAD \)

\( = 2(\angle ADC + \angle BAD) \) ......(3)

But in \( \triangle PAD \),

Ext. \( \angle APC = \angle PAD + \angle ADC \)

\( = \angle BAD + \angle ADC \) ........(4)

From (3) and (4),

\( \angle AOC + \angle BOD = 2\angle APC \)

Next part:

\( \angle BOD = 2\angle BAD = 2 \times 60° = 120° \)

and \( \angle BCD = \frac{1}{2} \) Re flex \( \{\angle BOD\} = \frac{1}{2}(360° - 120°) = 120° \)

Angle at the centre is double the angle at the circumference subtended by the same chord

\( \therefore \angle CBD + \angle CDB = 180° - 120° = 60° \)

By angle sum property of triangle CBD

Again, \( \angle OBD + \angle ODB = 180° - 120° = 60° \)

By angle sum property of triangle OBD

\( \therefore \angle OBD + \angle ODB = \angle CBD + \angle CDB \)

In simple words: We use the fact that the angle at the center is double the angle at the edge of the circle. This helps us prove both sides are equal to 60°.

📝 Teacher's Note: Draw a circle on the board. Show how angles at the center are always double the angles at the edge. This is the main rule we use in this proof.

🎯 Exam Tip: Always write "angle at center = 2 × angle at circumference" clearly. This is the key theorem. Also show all angle calculations step by step.

Question 25. In the figure given RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°

Calculate:
(i) ∠RNM;
(ii) ∠NRM.

[Diagram: This diagram shows a circle with RS as a diameter (a straight line through the center). NM is another line parallel to RS, and there is a 29° angle marked at MRS.]

Answer:

(i) Join RN and MS.

\( \angle RMS = 90° \) (Angle in a semicircle is a right angle)

\( \angle RSM = 90° - 29° = 61° \) (By angle sum property of triangle RMS)

\( \angle RNM = 180° - \angle RSM = 180° - 61° = 119° \) (Pair of opposite angles in a cyclic quadrilateral are supplementary)

(ii) Also, RS || NM

\( \angle NMR = \angle MRS = 29° \) (Alternate angles)

\( \angle NMS = 90° + 29° = 119° \)

Also, \( \angle NRS + \angle NMS = 180° \) (Pair of opposite angles in a cyclic quadrilateral are supplementary)

\( \Rightarrow \angle NRM + 29° + 119° = 180° \)

\( \Rightarrow \angle NRM = 180° - 148° \)

\( \therefore \angle NRM = 32° \)

In simple words: We use two main rules here. First, any angle in a semicircle is 90°. Second, opposite angles in a cyclic quadrilateral add up to 180°.

📝 Teacher's Note: Show students a semicircle and explain that any triangle with one side as diameter makes a 90° angle. This is very useful in circle problems.

🎯 Exam Tip: Write "angle in semicircle = 90°" and "opposite angles in cyclic quadrilateral = 180°" clearly. These are the two key rules for this problem.

Question 26. In the figure given alongside, AB || CD and O is the centre of the circle. If ∠ADC = 25°; find the angle AEB. Give reasons in support of your answer.

[Diagram: This diagram shows a circle with center O. AB and CD are parallel chords. There is a 25° angle marked, and point E appears to be where some lines meet.]

Answer:

Join AC and BD.

\( \angle CAD = 90° \) and \( \angle CBD = 90° \) (Angle in a semicircle is a right angle)

Also, AB || CD

\( \angle BAD = \angle ADC = 25° \) (Alternate angles)

\( \angle BAC = \angle BAD + \angle CAD = 25° + 90° = 115° \)

\( \angle ADB = 180° - 25° - \angle BAC = 180° - 25° - 115° = 40° \) (Pair of opposite angles in a cyclic quadrilateral are supplementary)

Also, \( \angle AEB = \angle ADB = 40° \) (Angles subtended by the same chord on the circle are equal)

In simple words: We use parallel lines to find equal alternate angles. Then we use the fact that angles made by the same chord are equal anywhere on the circle.

📝 Teacher's Note: Draw parallel lines and show how alternate angles are equal. Then show how angles subtended by the same chord are equal at any point on the circle.

🎯 Exam Tip: Write "alternate angles are equal" for parallel lines. Write "angles subtended by same chord are equal" for circle angles. These are key reasons.

Question 27. Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at C and D. Prove that AC is parallel to BD.

[Diagram: This diagram shows two circles that cross each other at points P and Q. Line APB goes through P and meets both circles. Line CQD goes through Q and meets both circles.]

Answer:

Join AC, PQ and BD.

ACQP is a cyclic quadrilateral

\( \angle CAP + \angle PQC = 180° \) ......(i) (Pair of opposite angles in a cyclic quadrilateral are supplementary)

PQDB is a cyclic quadrilateral

\( \angle PQD + \angle DBP = 180° \) ......(ii) (Pair of opposite angles in a cyclic quadrilateral are supplementary)

Again, \( \angle PQC + \angle PQD = 180° \) ......(iii) (CQD is a straight line)

Using (i), (ii) and (iii),

\( \angle CAP + \angle DBP = 180° \)

or \( \angle CAB + \angle DBA = 180° \)

We know, if a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel

\( \therefore AC || BD \)

In simple words: When two lines cut by a third line make interior angles that add to 180°, the two lines are parallel. We use circle properties to prove this.

📝 Teacher's Note: Draw two parallel lines cut by a third line. Show how interior angles on same side add to 180°. This is how we prove lines are parallel.

🎯 Exam Tip: Write "opposite angles in cyclic quadrilateral = 180°" first. Then write "interior angles on same side = 180° means lines are parallel". This completes the proof.

Question 28. ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.

[Diagram: This diagram shows two circles intersecting, with a cyclic quadrilateral ABCD. Lines AB and DC are extended to meet at point P outside the circles.]

Answer:

Join AC, PQ and BD.

ACQP is a cyclic quadrilateral

\( \angle CAP + \angle PQC = 180° \) ......(i) (Pair of opposite angles in a cyclic quadrilateral are supplementary)

PQDB is a cyclic quadrilateral

\( \angle PQD + \angle DBP = 180° \) ......(ii) (Pair of opposite angles in a cyclic quadrilateral are supplementary)

Again, \( \angle PQC + \angle PQD = 180° \) ......(iii) (CQD is a straight line)

Using (i), (ii) and (iii),

\( \angle CAP + \angle DBP = 180° \)

or \( \angle CAB + \angle DBA = 180° \)

We know, if a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel

\( \therefore AC || BD \)

In simple words: Since PA = PD, triangle PAD is equal-sided on two sides. This creates equal angles that help prove the lines are parallel using interior angle rules.

📝 Teacher's Note: When two sides of a triangle are equal, the angles opposite to them are also equal. Use this fact along with cyclic quadrilateral properties.

🎯 Exam Tip: Write "PA = PD given" first. Then use "angles opposite equal sides are equal" and "cyclic quadrilateral opposite angles = 180°". Show all steps clearly.

Question 29. AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find:
(i) ∠PRB
(ii) ∠PBR
(iii) ∠BPR.

[Diagram: Circle with diameter AB, points P and R on the circle, external point Q with straight lines APQ and RBQ. Angle measurements of 35° and 25° are marked in the diagram.]

Answer:
(i) ∠PRB = ∠PAB = 35°
Angles subtended by the same chord on the circle are equal.

(ii) ∠BPA = 90°
Angle in a semicircle is a right angle.
∠BPQ = 90°
∠PBR = ∠BQP + ∠BPQ = 25° + 90° = 115°
Exterior angle of a triangle is equal to the sum of pair of interior opposite angles.

(iii) ∠ABP = 90° - ∠BAP = 90° - 35° = 55°
∠ABR = ∠PBR - ∠ABP = 115° - 55° = 60°
∠APR = ∠ABR = 60°
Angles subtended by the same chord on the circle are equal.
∠BPR = 90° - ∠APR = 90° - 60° = 30°

In simple words: We use two main rules here. First, angles made by the same arc are equal. Second, any angle in a semicircle is 90°. We find each angle step by step using these rules.

📝 Teacher's Note: Draw a semicircle on the board and mark any point on it. Show students that the angle is always 90°. This is the most important property to remember for this type of problem.

🎯 Exam Tip: Always write the reason after each step. Write "angle in semicircle = 90°" and "angles subtended by same chord are equal". These exact phrases get you marks.

Question 30. In the given figure, SP is the bisector of angle RPT and PQRS is a cyclic quadrilateral. Prove that: SQ = SR.

[Diagram: Circle with cyclic quadrilateral PQRS, external point T with straight line QPT, and SP bisecting angle RPT.]

Answer:
PQRS is a cyclic quadrilateral
∠QRS + ∠QPS = 180° ... (i)
(Pair of opposite angles in a cyclic quadrilateral are supplementary)

Also, ∠QPS + ∠SPT = 180° ... (ii)
(Straight line QPT)

From (i) and (ii),
∠QRS = ∠SPT ... (iii)

Also, ∠RQS = ∠RPS ... (iv)
(Angles subtended by the same chord on the circle are equal)

and ∠RPS = ∠SPT (PS bisects ∠RPT) ... (v)

From (iii), (iv) and (v),
∠QRS = ∠RQS
∴ SQ = SR

In simple words: We show that two angles in triangle QRS are equal. When two angles in a triangle are equal, the sides opposite to them are also equal. So SQ = SR.

📝 Teacher's Note: Remind students that in any triangle, equal angles have equal opposite sides. This is called isosceles triangle property. Draw a simple isosceles triangle to show this.

🎯 Exam Tip: Write all the angle relationships step by step. Always mention "cyclic quadrilateral" and "angle bisector" properties clearly. End with "equal angles, so equal sides".

Question 31. In the figure, O is the centre of the circle, ∠AOE = 150°, DAO = 51°. Calculate the sizes of the angles CEB and OCE.

[Diagram: Circle with center O, points A, B, C, D, E on the circle, with angle AOE = 150° and angle DAO = 51° marked.]

Answer:
Step 1: Find ∠ADE
∠ADE = \( \frac{1}{2} \) Reflex(∠AOE) = \( \frac{1}{2}(360° - 150°) = 105° \)
Angle at the centre is double the angle at the circumference subtended by the same chord.

Step 2: Find ∠BED
∠DAB + ∠BED = 180°
Pair of opposite angles in a cyclic quadrilateral are supplementary.
∠BED = 180° - 51° = 129°

Step 3: Find ∠CEB
∠CEB = 180° - ∠BED = 180° - 129° = 51°
(Straight line)

Step 4: Find ∠OCE
By angle sum property of △ADC,
∠OCE = 180° - 51° - 105° = 24°

In simple words: We use the rule that center angle is double the circumference angle. Then we use properties of cyclic quadrilaterals to find other angles step by step.

📝 Teacher's Note: Show students with a paper circle that the center angle is always twice the angle at the edge. Cut out a paper circle and demonstrate with a pencil compass.

🎯 Exam Tip: Always write "center angle = 2 × circumference angle" as your first step. Show all working clearly. Don't skip the reflex angle calculation.

Question 32. In the figure, P and Q are the centres of two circles intersecting at B and C. ACD is a straight line. Calculate the numerical value of x.

[Diagram: Two intersecting circles with centers P and Q, intersecting at points B and C, with straight line ACD and angle x marked at Q, and 150° marked at P.]

Answer:
Step 1: Find ∠ACB
∠ACB = \( \frac{1}{2} \) ∠APB = \( \frac{1}{2} × 150° = 75° \)
Angle at the centre is double the angle at the circumference subtended by the same chord.

Step 2: Find ∠BCD
∠ACB + ∠BCD = 180° (Straight line)
∠BCD = 180° - 75° = 105°

Step 3: Find x
Also, ∠BCD = \( \frac{1}{2} \) Reflex∠BQD = \( \frac{1}{2}(360° - x) \)
Angle at the centre is double the angle at the circumference subtended by the same chord.
105° = 180° - \( \frac{x}{2} \)
x = 2(180° - 105°) = 2 × 75° = 150°

In simple words: We find the angle at C first using the center-circumference rule. Then we use the fact that ACD is a straight line. Finally we apply the same rule to the other circle to find x.

📝 Teacher's Note: Draw two overlapping circles on the board. Show students how the same point C can be on both circles, so we can use the center-circumference rule for both circles.

🎯 Exam Tip: Write "center angle = 2 × circumference angle" for both circles. Use "straight line = 180°" clearly. Show the reflex angle calculation step by step.

Question 33. The figure shows two circles which intersect at A and B. The centre of the smaller circle is O and lies on the circumference of the larger circle. Given that ∠APB = a°. Calculate, in terms of a°, the value of:
(i) obtuse ∠AOB
(ii) ∠ACB
(iii) ∠ADB.
Give reasons for your answers clearly.

[Diagram: Two intersecting circles where the smaller circle's center O lies on the larger circle's circumference, with points A, B, C, D, P marked and angle APB = a°.]

Answer:
(i) Obtuse ∠AOB = (180° - a°)
Since O lies on the circumference of the larger circle, ∠APB and ∠AOB are angles subtended by chord AB.
∠APB + ∠AOB = 180° (angles on opposite sides of chord in a circle)
Therefore, ∠AOB = 180° - a°

(ii) ∠ACB = \( \frac{a°}{2} \)
In the smaller circle with center O:
Central angle ∠AOB = 180° - a°
∠ACB = \( \frac{1}{2} \) × (180° - a°) = 90° - \( \frac{a°}{2} \)
But ∠ACB and ∠APB are on the same side of chord AB in the larger circle
∠ACB = \( \frac{a°}{2} \)

(iii) ∠ADB = (90° - \( \frac{a°}{2} \))
∠ADB is the angle subtended by chord AB at point D on the smaller circle.
∠ADB = \( \frac{1}{2} \) × central angle ∠AOB = \( \frac{1}{2} \) × (180° - a°) = 90° - \( \frac{a°}{2} \)

In simple words: We use two rules here. First, when center of one circle lies on another circle, special angle relationships form. Second, center angle is always double the circumference angle in any circle.

📝 Teacher's Note: This is a complex problem. Draw the diagram carefully and show students how one circle's center lies on another circle. This creates special angle relationships that are not obvious.

🎯 Exam Tip: Always state which circle you are using for each calculation. Write "in the larger circle" or "in the smaller circle" clearly. Show all angle relationships step by step with reasons.

Question 34. In the given figure, O is the centre of the circle and ∠ABC = 55°. Calculate the values of x and y.
Answer:

[Diagram: Circle with center O, showing points A, B, C, D, and P with various angles marked including x and y]

Step 1: Find angle AOB
∠AOB = 2∠APB = 2a° (Angle at the centre is double the angle at the circumference subtended by the same chord)

Step 2: Use cyclic quadrilateral property
OABC is a cyclic quadrilateral
∠AOB + ∠ACB = 180° (Pair of opposite angles in a cyclic quadrilateral are supplementary)
⇒ ∠ACB = 180° - 2a°

Step 3: Find angles using chord properties
Join AB.
∠ADB = ∠ACB = 180° - 2a° (Angles subtended by the same arc on the circle are equal)

In simple words: We use the rule that angle at center is double the angle at circumference. Then we use cyclic quadrilateral properties where opposite angles add to 180°.

📝 Teacher's Note: Draw a pizza slice to show center angle. The angle you make at the center when you cut the slice is always double the angle someone sitting on the edge of the pizza sees.

🎯 Exam Tip: Always state "angle at center is double angle at circumference" clearly. Write the cyclic quadrilateral property that opposite angles are supplementary.

Question 35. In the given figure, A is the centre of the circle, ABCD is a parallelogram and CDE is a straight line. Prove that ∠BCD = 2∠ABE
Answer:

[Diagram: Circle with center A, parallelogram ABCD inscribed, with point E on the circle forming straight line CDE]

∠AOC = 2∠ABC = 2 × 55°
(Angle at the centre is double the angle at the circumference subtended by the same chord)

∴ x = 110°
ABCD is a cyclic quadrilateral
∴ ∠ADC + ∠ABC = 180°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)

⇒ y = 180° - 55° = 125°

In simple words: The center angle is always double the edge angle for the same arc. In a cyclic quadrilateral, opposite angles always add up to 180°.

📝 Teacher's Note: Show students that when four points lie on a circle, opposite angles of the quadrilateral always add to 180°. This is a key property.

🎯 Exam Tip: Write x = 110° and y = 125° clearly as final answers. Always mention which circle theorem you are using.

Question 36. ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°; calculate: (i) ∠DAB, (ii) ∠BDC.
Answer:

[Diagram: Circle with diameter AB, cyclic quadrilateral ABCD with AB parallel to DC, point E on circle]

(i) ∠BAD = 2∠BED
(Angle at the centre is double the angle at the circumference subtended by the same chord)

and ∠BED = ∠ABE (Alternate angles)
∴ ∠BAD = 2∠ABE --- (i)

ABCD is a parallelogram
∴ ∠BAD = ∠BCD --- (ii)
(Opposite angles in a parallelogram are equal)

From (i) and (ii),
∠BCD = 2∠ABE.

In simple words: We use the center-circumference angle rule and properties of parallelograms. The proof shows one angle equals double another angle.

📝 Teacher's Note: Explain that when AB is parallel to DC, we get alternate angles. Combine this with the center angle theorem to get the result.

🎯 Exam Tip: Write "∠BCD = 2∠ABE" as the final statement to prove. Show each step clearly with reasons in brackets.

Question 37. In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°; calculate: (i) ∠EBA, (ii) ∠BCD.
Answer:

[Diagram: Circle with diameter AB, chord ED parallel to AB, various angles marked]

(i) ∠DAB = ∠BED = 65°
(Angles subtended by the same chord on the circle are equal)

(ii) ∠ADB = 90°
(Angle in a semicircle is a right angle)

∴ ∠ABD = 90° - ∠DAB = 90° - 65° = 25°
AB ∥ DC

∴ ∠BDC = ∠ABD = 25° (Alternate angles)

In simple words: When AB is a diameter, any angle in the semicircle is 90°. Parallel lines give us equal alternate angles.

📝 Teacher's Note: Emphasize that diameter always makes 90° angle at any point on the circle. This is a very important theorem.

🎯 Exam Tip: Always write "angle in semicircle = 90°" when AB is diameter. Then use parallel line properties for alternate angles.

Question 38. In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°; calculate: (i) ∠DAB, (ii) ∠DBA, (iii) ∠DBC, (iv) ∠ADC. Also, show that the △AOD is an equilateral triangle.
Answer:

[Diagram: Circle with center O, diameter AB, point D and C on circle, with DO parallel to CB]

(i) ∠AEB = 90°
(Angle in a semicircle is a right angle)

Therefore ∠EBA = 90° - ∠EAB = 90° - 63° = 27°

(ii) AB ∥ ED
Therefore ∠DEB = ∠EBA = 27° (Alternate angles)

Therefore BCDE is a cyclic quadrilateral
Therefore ∠DEB + ∠BCD = 180°

[Pair of opposite angles in a cyclic quadrilateral are supplementary]

Therefore ∠BCD = 180° - 27° = 153°

In simple words: We use the 90° angle rule for diameter, then alternate angles for parallel lines, and finally cyclic quadrilateral properties.

📝 Teacher's Note: Show that when lines are parallel, alternate angles are equal. Combine this with semicircle angle = 90° and cyclic quadrilateral properties.

🎯 Exam Tip: Write each angle calculation step by step. State which theorem you use for each step. Final answers: ∠EBA = 27°, ∠BCD = 153°.

[Diagram: Circle with points A, D, C, B marked on the circumference, with lines connecting them to form a cyclic quadrilateral ABCD, and point O at the center]

Solution:
Answer:
(i) ABCD is a cyclic quadrilateral
∴ \( \angle DCB + \angle DAB = 180° \)
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
\( \Rightarrow \angle DAB = 180° - 120° = 60° \)
(ii) \( \angle ADB = 90° \)
(Angle in a semicircle is a right angle)
∴ \( \angle DBA = 90° - \angle DAB = 90° - 60° = 30° \)
(iii) OD = OB
∴ \( \angle ODB = \angle OBD \)
or \( \angle ABD = 30° \)
Also, AB ∥ ED
∴ \( \angle DBC = \angle ODB = 30° \) (Alternate angles)
(iv) \( \angle ABD + \angle DBC = 30° + 30° = 60° \)
\( \Rightarrow \angle ABC = 60° \)
In cyclic quadrilateral ABCD,
\( \angle ADC + \angle ABC = 180° \)
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
\( \Rightarrow \angle ADC = 180° - 60° = 120° \)
In △AOD, OA=OD (radii of the same circle)
\( \angle AOD = \angle DAO \) or \( \angle DAB=60° \) [proved in (i)]
\( \Rightarrow \angle AOD = 60° \)
\( \angle ADO = \angle AOD = \angle DAO = 60° \)
∴ △AOD is an equilateral triangle.

📝 Teacher's Note: Draw a circle and mark the points clearly. Show that opposite angles add up to 180°. This is the most important property of cyclic quadrilaterals.

🎯 Exam Tip: Always write "opposite angles are supplementary" when dealing with cyclic quadrilaterals. Also mention "angle in semicircle is 90°" when needed.

Question 39. In the given figure, I is the in centre of the △ ABC. BI when produced meets the circumcircle of △ ABC at D. Given ∠BAC = 55° and ∠ ACB = 65°, calculate:
(i) ∠DCA,
(ii) ∠ DAC,
(iii) ∠DCI,
(iv) ∠AIC.

[Diagram: Triangle ABC inscribed in a circle with incenter I, and BI extended to meet the circle at D]

Solution:
Answer:
Join IA, IC and CD.

(i) IB is the bisector of ∠ABC
\( \Rightarrow \angle ABD = \frac{1}{2} \angle ABC = \frac{1}{2}(180° - 65° - 55°) = 30° \)
∠ DCA = ∠ABD = 30°
(Angle in the same segment)

(ii) ∠DAC = ∠CBD = 30°
(Angle in the same segment)

(iii) \( \angle ACI = \frac{1}{2} \angle ACB = \frac{1}{2} × 65° = 32.5° \)
(CI is the angular bisector of ∠ACB)
∴ ∠DCI = ∠DCA + ∠ACI = 30° + 32.5° = 62.5°

(iv) \( \angle IAC = \frac{1}{2} \angle BAC = \frac{1}{2} × 55° = 27.5° \)
(AI is the angular bisector of ∠BAC)
∴ ∠AIC = 180° - ∠IAC - ∠ICA = 180° - 27.5° - 32.5° = 120°

📝 Teacher's Note: The incenter is where all angle bisectors meet. Remember that angles in the same segment of a circle are equal. This helps solve many parts quickly.

🎯 Exam Tip: Write "I is the incenter" and "angles in same segment are equal" clearly. These are key points that get you marks.

Question 40. A triangle ABC is inscribed in a circle. The bisectors of angles BAC, ABC and ACB meet the circumcircle of the triangle at points P, Q and R respectively. Prove that:
(i) ∠ABC = 2 ∠APQ
(ii) ∠ACB = 2 ∠APR
(iii) ∠QPR = 90° - ∠BAC

[Diagram: Triangle ABC inscribed in a circle with angle bisectors meeting the circle at points P, Q, and R]

Solution:
Answer:
Join PQ and PR.

(i) BQ is the bisector of ∠ABC
\( \Rightarrow \angle ABQ = \frac{1}{2} \angle ABC \)
Also, ∠ APQ = ∠ABQ
(Angle in the same segment)
∴ ∠ABC = 2 ∠APQ

(ii) CR is the bisector of ∠ACB
\( \Rightarrow \angle ACR = \frac{1}{2} \angle ACB \)
Also, ∠ ACR = ∠APR
(Angle in the same segment)
∴ ∠ACB = 2 ∠APR

(iii) Adding (i) and (ii),
we get
∠ABC + ∠ACB = 2(∠APR + ∠APQ) = 2∠QPR
\( \Rightarrow 180° - \angle BAC = 2∠QPR \)
\( \Rightarrow \angle QPR = 90° - \frac{1}{2} \angle BAC \)

📝 Teacher's Note: This is about angle bisectors meeting the circumcircle. The key idea is that angles in the same segment are equal. Use this property step by step.

🎯 Exam Tip: Write each step clearly. Show that angles are in the same segment. Then use the angle bisector property. Don't skip the "angles in same segment" statement.

Question 40 (old). The sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E; the sides DA and CB are produced to meet at F. If ∠BEC = 42° and ∠BAD = 98°; calculate:
(i) ∠AFB,
(ii) ∠ADC.

[Diagram: Cyclic quadrilateral ABCD with sides extended to meet at points E and F, showing angles 98° and 42°]

Solution:
Answer:
By angle sum property of △ADE,
∠ADC = 180° - 98° - 42° = 40°

Also, ∠ADC + ∠ABC = 180°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
∴ ∠ ABC = 180° - 40° = 140°

Also, ∠BAF = 180° - ∠BAD = 180° - 98° = 82°
∴ ∠ABC = ∠AFB + ∠BAF
(Exterior angle of a △ is equal to the sum of pair of interior opposite angles)
\( \Rightarrow \angle AFB = 140° - 82° = 58° \)

Thus, ∠AFB = 58° and ∠ADC = 40°

📝 Teacher's Note: When sides of a cyclic quadrilateral are extended, use the exterior angle property. Remember that opposite angles in a cyclic quadrilateral add up to 180°.

🎯 Exam Tip: Write "opposite angles are supplementary" and "exterior angle property" clearly. These are the main properties you need to use here.

Question 41. Calculate the angles x, y and z if: \( \frac{x}{3} = \frac{y}{4} = \frac{z}{5} \)

[Diagram: This diagram shows a circle with points A, B, C, D, P, Q. There is a quadrilateral ABCD inscribed in the circle with point C at the bottom. Lines extend from B to P and from D to Q forming a triangle PBQ outside the circle. Angles are marked as x at C, y at P, and z at Q.]

Answer:
Step 1: Let x = 3k, y = 4k and z = 5k (where k is a constant)

Step 2: Find the value of k using angle properties.
∠ADB = x + z = 8k and ∠ABC = x + y = 7k
(Exterior angle of a triangle is equal to the sum of pair of interior opposite angles)

Step 3: Use the property of cyclic quadrilateral.
Also, ∠ABC + ∠ADC = 180°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)

Step 4: Calculate k.
8k + 7k = 180°
15k = 180°
k = \( \frac{180°}{15} = 12° \)

Step 5: Find the angles.
x = 3 × 12° = 36°
y = 4 × 12° = 48°
z = 5 × 12° = 60°

In simple words: We used the fact that angles are in ratio 3:4:5. In a cyclic quadrilateral, opposite angles add up to 180°. This helped us find the common factor k = 12°.

📝 Teacher's Note: Draw a cyclic quadrilateral on the board. Show students that opposite angles always add to 180°. This is the key property to remember.

🎯 Exam Tip: Always write the ratio clearly first. Then use the cyclic quadrilateral property that opposite angles are supplementary (add to 180°).

Question 42. In the given figure, AB = AC = CD and ∠ADC = 38°. Calculate:
(i) Angle ABC
(ii) Angle BEC.

[Diagram: This diagram shows a circle with points A, B, C, D, E. There are triangles and chords formed, with AB = AC = CD marked as equal lengths. Point E appears to be outside the main triangle configuration.]

Answer:
(i) Finding Angle ABC:
AC = CD
∴ ∠CAD = ∠CDA = 38°
∴ ∠ACD = 180° - 2 × 38° = 104°
∴ ∠ACB = 180° - 104° = 76° (Straight line)
Also, AB = AC
∴ ∠ABC = ∠ACB = 76°

(ii) Finding Angle BEC:
By angle sum property,
∠BAC = 180° - 2 × 76° = 28°
∴ ∠BEC = ∠BAC = 28°
(Angles in the same chord)

In simple words: When two sides are equal in a triangle, the angles opposite to them are also equal. We used this property twice - once for triangle ACD and once for triangle ABC.

📝 Teacher's Note: Show students that equal sides create equal angles. Use a compass to draw equal sides and measure the opposite angles. They will see they are equal.

🎯 Exam Tip: Always write "equal sides, equal angles" when AB = AC. Then use the property that angles in the same chord are equal for part (ii).

Question 43. In the given figure, AC is the diameter of circle, centre O. Chord BD is perpendicular to AC. Write down the angles p, and r in terms of x.

[Diagram: This diagram shows a circle with center O. AC is the diameter (horizontal line). BD is a chord perpendicular to AC, intersecting it inside the circle. Angles are marked as p, x, r, and q at various positions.]

Answer:
Step 1: Use the property of angles at center and circumference.
∠AOB = 2∠ACB = 2∠ADB
(Angle at the centre is double the angle at the circumference subtended by the same chord)

\( \Rightarrow \) x = 2q and ∠ADB = \( \frac{x}{2} \)
∴ q = \( \frac{x}{2} \)

Step 2: Find angle r.
Also, ∠ADC = 90°
(Angle in a semicircle)
\( \Rightarrow \) r + \( \frac{x}{2} \) = 90°
\( \Rightarrow \) r = 90° - \( \frac{x}{2} \)

Step 3: Find angle p.
Again, ∠DAC = ∠DBC
(Angle in the same segment)
\( \Rightarrow \) p = 90° - q
\( \Rightarrow \) p = 90° - \( \frac{x}{2} \)

Therefore:
p = 90° - \( \frac{x}{2} \)
r = 90° - \( \frac{x}{2} \)

In simple words: When AC is a diameter, any angle on the circle is 90°. We used this property along with the fact that the angle at center is twice the angle at circumference.

📝 Teacher's Note: Draw a semicircle and show students that any triangle with the diameter as base has a 90° angle. This is Thales' theorem.

🎯 Exam Tip: Remember: angle in semicircle = 90°. Also, angle at center = 2 × angle at circumference. These are the two key properties needed.

Question 44. In the given figure, AC is the diameter of circle, centre O. CD and BE are parallel. Angle AOB = 80° and angle ACE = 10°. Calculate:
(i) Angle BEC;
(ii) Angle BCD;
(iii) Angle CED.

[Diagram: This diagram shows a circle with center O. AC is the diameter. Points B, D, E are on the circle with CD and BE parallel to each other. Various angles are marked including 80° and 10°.]

Answer:
(i) Finding Angle BEC:
∠BOC = 180° - 80° = 100° (Straight line)
and ∠BOC = 2∠BEC
(Angle at the centre is double the angle at the circumference subtended by the same chord)
\( \Rightarrow \) ∠BEC = \( \frac{100°}{2} = 50° \)

(ii) Finding Angle BCD:
DC || EB
∴ ∠DCE = ∠BEC = 50° (Alternate angles)
∴ ∠AOB = 80°
\( \Rightarrow \) ∠ACB = \( \frac{1}{2} \)∠AOB = 40°
(Angle at the centre is double the angle at the circumference subtended by the same chord)

We have,
∠BCD = ∠ACB + ∠ACE + ∠DCE = 40° + 10° + 50° = 100°

(iii) Finding Angle CED:
∠BED = 180° - ∠BCD = 180° - 100° = 80°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
\( \Rightarrow \) ∠CED + 50° = 80°
\( \Rightarrow \) ∠CED = 30°

In simple words: We used parallel lines (alternate angles are equal) and the circle properties. In a cyclic quadrilateral, opposite angles add up to 180°.

📝 Teacher's Note: Show students how parallel lines create equal alternate angles. Then combine this with circle angle properties. Draw the parallel lines clearly.

🎯 Exam Tip: Mark the parallel lines first. Write "alternate angles" when using parallel line properties. For circles, always use "angle at center = 2 × angle at circumference".

Question 45. In the given figure, AE is the diameter of circle. Write down the numerical value of ∠ABC + ∠CDE. Give reasons for your answer.

[Diagram: This diagram shows a circle with AE as the diameter. Points B, C, D are on the circle forming angles ABC and CDE. There is a center point O marked.]

Answer:
Join centre O and C and EC.

∠AOC = \( \frac{180°}{2} = 90° \)

and ∠AOC = 2∠AEC
(Angle at the centre is double the angle at the circumference subtended by the same chord)

Since AE is the diameter, all angles in the semicircle are 90°.
∠ABC and ∠CDE are angles in the same semicircle.

For any cyclic quadrilateral ABCDE inscribed in a semicircle:
∠ABC + ∠CDE = 180°

This is because they are angles subtending the same arc AE (which is a semicircle).

Therefore: ∠ABC + ∠CDE = 180°

In simple words: When AE is a diameter, the angles ABC and CDE are on opposite sides of the circle. They always add up to 180° because they are in a cyclic quadrilateral.

📝 Teacher's Note: Draw a circle with diameter. Mark any two points and show that angles from these points to the diameter always add to 180°. This is a key circle property.

🎯 Exam Tip: When you see a diameter, remember that opposite angles in the resulting cyclic quadrilateral add to 180°. Write this property clearly in your answer.

Question 46. In the given figure, AOC is a diameter and AC is parallel to ED. If ∠CBE = 64°, calculate ∠DEC.
Answer:
Join AB.
∠ABC = 90°
(Angle in a semicircle)
∴ ∠ABE = 90° - 64° = 26°
Now, ∠ABE = ∠ACE = 26°
(Angle in the same segment)
Also, AC || ED
∴ ∠DEC = ∠ACE = 26° (Alternate angles)
In simple words: We use the fact that angles in the same part of a circle are equal. Since AC and ED are parallel lines, we get alternate angles that are equal.

[Diagram: This diagram shows a circle with diameter AOC, points B, D, E on the circle, and parallel lines AC and ED, with angle CBE marked as 64°.]

📝 Teacher's Note: First find the angle in semicircle (90°). Then use same segment rule. Finally use parallel lines to get alternate angles. Show each step clearly.

🎯 Exam Tip: Always write "angle in semicircle = 90°" and "alternate angles are equal" clearly. These are key phrases examiners look for.

Question 47. Use the given figure to find
(i) ∠BAD
(ii) ∠DQB.
Answer:
(i) By angle sum property of ΔADP,
∠BAD = 180° - 85° - 40° = 55°
(ii) ∠ABC = 180° - ∠ADC = 180° - 85° = 95°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
By angle sum property,
∠AQB = 180° - 95° - 55°
⇒ ∠DQB = 30°
In simple words: In a cyclic quadrilateral, opposite angles add up to 180°. We use this rule and triangle angle sum property to find the required angles.

[Diagram: This diagram shows a circle with points A, B, C, D on the circle, external point P connected to C, and point Q, with angles 85° and 40° marked.]

📝 Teacher's Note: Teach students that in any cyclic quadrilateral, opposite angles are supplementary (add to 180°). This is a very important circle theorem.

🎯 Exam Tip: Write "opposite angles in cyclic quadrilateral are supplementary" clearly. Always show your angle sum calculations step by step.

Question 48. In the given figure, AOB is a diameter and DC is parallel to AB. If ∠CAB = x°; find (in terms of x) the values of:
(i) ∠COB
(ii) ∠DOC
(iii) ∠DAC
(iv) ∠ADC.
Answer:
(i) ∠COB = 2∠CAB = 2x
(Angle at the centre is double the angle at the circumference subtended by the same chord)
(ii) ∠OCD = ∠COB = 2x (Alternate angles)
In ΔOCD, OC = OD
∴ ∠ODC = ∠OCD = 2x
By angle sum property of ΔOCD,
∠DOC = 180° - 2x - 2x = 180° - 4x
(iii) ∠DAC = \( \frac{1}{2} \)∠DOC = \( \frac{1}{2} \)(180° - 4x) = 90° - 2x
(Angle at the centre is double the angle at the circumference subtended by the same chord)
(iv) DC || AO
∴ ∠ACD = ∠OAC = x (Alternate angles)
By angle sum property,
∠ADC = 180° - ∠DAC - ∠ACD = 180° - (90° - 2x) - x = 90° + x
In simple words: We use the rule that centre angle is double the circumference angle. We also use properties of parallel lines and isosceles triangles.

[Diagram: This diagram shows a circle with diameter AOB, points C and D on the circle, with DC parallel to AB, and angle CAB marked as x°.]

📝 Teacher's Note: This question uses multiple circle theorems together. Teach students to identify which theorem applies at each step. Centre angle = 2 × circumference angle is key.

🎯 Exam Tip: Always mention "angle at centre is double angle at circumference" and "alternate angles are equal" when using these properties. Show all algebraic steps clearly.

Question 49. In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 130°. Find:
(i) ∠DAB
(ii) ∠DBA
Answer:
i. ABCD is a cyclic quadrilateral
∴ ∠DAB = 180° - ∠DCB
= 180° - 130°
= 50°
ii. In ΔADB,
∠DAB + ∠ADB + ∠DBA = 180°
⇒ 50° + 90° + ∠DBA = 180°
⇒ ∠DBA = 40°
In simple words: Opposite angles in a cyclic quadrilateral add to 180°. Also, angle in semicircle is always 90°.

[Diagram: This diagram shows a circle with diameter AB, points C and D on the circle, with angle BCD marked as 130°.]

📝 Teacher's Note: First use the cyclic quadrilateral property for opposite angles. Then remember that any angle in a semicircle (with diameter as one side) is 90°.

🎯 Exam Tip: Write "opposite angles in cyclic quadrilateral are supplementary" and "angle in semicircle = 90°". These exact phrases get you marks.

Question 50. In the given figure, PQ is the diameter of the circle whose centre is O. Given ∠ROS = 42°; calculate ∠RTS.
Answer:
Join PS.
∠PSQ = 90°
(Angle in a semicircle)
Also, ∠SPR = \( \frac{1}{2} \)∠ROS
(Angle at the centre is double the angle at the circumference subtended by the same chord)
⇒ ∠SPT = \( \frac{1}{2} \) × 42° = 21°
∴ In right triangle PST,
∠PTS = 90° - ∠SPT
⇒ ∠RTS = 90° - 21° = 69°
In simple words: Angle at centre is double the angle at circumference. Angle in semicircle is 90°. We use these two facts to solve this problem.

[Diagram: This diagram shows a circle with diameter PQ, centre O, points R, S, T on the circle, with angle ROS marked as 42°.]

📝 Teacher's Note: Students often forget to join the required line (PS here). Always draw the complete diagram first. Centre angle = 2 × circumference angle is crucial.

🎯 Exam Tip: Write "Join PS" at the start. Then write "angle in semicircle = 90°" and "angle at centre = 2 × angle at circumference". Show each step clearly.

Question 51. In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°; calculate
(i) ∠RPQ
(ii) ∠STP.
Answer:
(i) In ΔPQR,
∠PRQ = 90° (Angle in semicircle)
∠RPQ = 180° - 90° - 58° = 32°
(ii) ∠SRP = ∠RPQ = 32° (Alternate angles, as SR || PQ)
∠STP = ∠SRP = 32° (Angles in same segment)
In simple words: Angle in semicircle is 90°. Parallel lines give alternate angles. Angles in same part of circle are equal.

[Diagram: This diagram shows a circle with diameter PQ, chord SR parallel to PQ, point T on the circle, with angle PQR marked as 58°.]

📝 Teacher's Note: This combines three important theorems: angle in semicircle, alternate angles with parallel lines, and angles in same segment. Teach students to identify each step.

🎯 Exam Tip: Always mention "angle in semicircle = 90°", "alternate angles are equal", and "angles in same segment are equal" when using these properties.

Question 52. AB is the diameter of the circle with centre O. OD is parallel to BC and ∠AOD = 60°; calculate the numerical values of:
(i) ∠ABD,
(ii) ∠DBC,
(iii) ∠ADC.
Answer:
[Diagram: This diagram shows a circle with centre O, where AB is the diameter, and points C and D are on the circle with OD parallel to BC.]

Join BD.

(i) ∠ABD = \( \frac{1}{2} \) ∠AOD = \( \frac{1}{2} \times 60° = 30° \)
This is because the angle at the centre is double the angle at the circumference subtended by the same chord.

(ii) ∠BDA = 90°
This is because any angle in a semicircle is 90°.

Also, triangle OAD is equilateral because ∠AOD = 60°.
Therefore, ∠ODB = 90° - ∠ODA = 90° - 60° = 30°

Also, OD || BC
Therefore, ∠DBC = ∠ODB = 30° (alternate angles)

(iii) ∠ABC = ∠ABD + ∠DBC = 30° + 30° = 60°
In cyclic quadrilateral ABCD,
∠ADC = 180° - ∠ABC = 180° - 60° = 120°
This is because opposite angles in a cyclic quadrilateral are supplementary.

In simple words: We use circle rules. The angle at the centre is twice the angle at the edge. Angles in a semicircle are 90°. Opposite angles in a four-sided shape inside a circle add up to 180°.

📝 Teacher's Note: Show students how to identify centre angles and circumference angles. Draw many examples on the board. Students often forget that AB is a diameter, so angles on the semicircle are 90°.

🎯 Exam Tip: Always write the rule you use. Write "angle at centre is double angle at circumference" and "angle in semicircle is 90°". These key phrases get you marks.

Question 53. In the given figure, the centre of the small circle lies on the circumference of the bigger circle. If ∠APB = 75° and ∠BCD = 40°; find:
(i) ∠AOB,
(ii) ∠ACB,
(iii) ∠ABD,
(iv) ∠ADB.
Answer:
[Diagram: This diagram shows two intersecting circles where the centre of the small circle lies on the circumference of the bigger circle.]

Join AB and AD.

(i) ∠AOB = 2∠APB = 2 × 75° = 150°
The angle at the centre is double the angle at the circumference subtended by the same chord.

(ii) In cyclic quadrilateral AOBC,
∠ACB = 180° - ∠AOB = 180° - 150° = 30°
Opposite angles in a cyclic quadrilateral are supplementary.

(iii) In cyclic quadrilateral ABDC,
∠ABD = 180° - ∠ACD = 180° - (40° + 30°) = 110°
Opposite angles in a cyclic quadrilateral are supplementary.

(iv) In cyclic quadrilateral AOBD,
∠ADB = 180° - ∠AOB = 180° - 150° = 30°
Opposite angles in a cyclic quadrilateral are supplementary.

In simple words: We use two main rules. First, the angle at the centre is twice the angle at the edge. Second, in a four-sided shape inside a circle, opposite angles add up to 180°.

📝 Teacher's Note: Draw two intersecting circles on the board. Show students how to identify which quadrilaterals are cyclic. Point out that not all four-sided shapes are cyclic quadrilaterals.

🎯 Exam Tip: Always state which quadrilateral you are using. Write "In cyclic quadrilateral ABCD" clearly. Also mention the property: "opposite angles are supplementary".

Question 54. In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°; find:
(i) ∠BCD,
(ii) ∠ACB.
Hence, show that AC is a diameter.
Answer:
[Diagram: This diagram shows a circle with points A, B, C, D on the circumference forming a cyclic quadrilateral.]

(i) In cyclic quadrilateral ABCD,
∠BCD = 180° - ∠BAD = 180° - 65° = 115°
Opposite angles in a cyclic quadrilateral are supplementary.

(ii) By angle sum property of triangle ABD,
∠ADB = 180° - 65° - 70° = 45°

Again, ∠ACB = ∠ADB = 45°
This is because angles in the same segment are equal.

Therefore, ∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°
Hence, AC is a semicircle.
Since angle in a semicircle is a right angle, AC must be a diameter.

In simple words: When we get 90° as an angle on the circle, it means the line across is a diameter. This is because angles in a semicircle are always 90°.

📝 Teacher's Note: Show students a semicircle and mark any point on it. The angle is always 90°. This is the key to identifying diameters. Students should remember this rule.

🎯 Exam Tip: When you get 90° in a circle problem, immediately check if it proves a diameter. Write "angle in semicircle is 90°, therefore AC is diameter". This gets full marks.

Question 55. In a cyclic quadrilateral ABCD, ∠A : ∠C = 3 : 1 and ∠B : ∠D = 1 : 5; find each angle of the quadrilateral.
Answer:

Let ∠A and ∠C be 3x and x respectively.
In cyclic quadrilateral ABCD,
∠A + ∠C = 180° (opposite angles are supplementary)
⇒ 3x + x = 180°
⇒ x = \( \frac{180°}{4} \) = 45°

Therefore, ∠A = 135° and ∠C = 45°

Let the measure of ∠B and ∠D be y and 5y respectively.
In cyclic quadrilateral ABCD,
∠B + ∠D = 180° (opposite angles are supplementary)
⇒ y + 5y = 180°
⇒ y = \( \frac{180°}{6} \) = 30°

Therefore, ∠B = 30° and ∠D = 150°

In simple words: In a four-sided shape inside a circle, opposite angles always add up to 180°. We use this rule with the given ratios to find all angles.

📝 Teacher's Note: Teach students to always let the ratio parts be variables. If ratio is 3:1, use 3x and x. This makes the algebra much easier to solve.

🎯 Exam Tip: Always write "opposite angles in cyclic quadrilateral are supplementary". Then set up the equation with ratios. Show all working steps clearly for full marks.

Question 56. The given figure shows a circle with centre O and ∠ABP = 42°. Calculate the measure of
(i) ∠PQB
(ii) ∠QPB + ∠PBQ
Answer:
[Diagram: This diagram shows a circle with centre O, with points A, B, P, Q on or related to the circle.]

(i) ∠PQB = ∠ABP = 42°
This is because angles in the same segment are equal.

(ii) In triangle PQB,
∠QPB + ∠PBQ + ∠PQB = 180° (angle sum property)
⇒ ∠QPB + ∠PBQ = 180° - 42° = 138°

In simple words: Angles that look at the same part of a circle are equal. In any triangle, all three angles add up to 180°.

📝 Teacher's Note: Draw several points on a circle looking at the same arc. Show that all these angles are equal. This is the "angles in same segment" rule that students must remember.

🎯 Exam Tip: Write "angles in same segment are equal" and "angle sum in triangle is 180°". These are the key statements that get you marks. Always show the triangle you are using.

Question 57. In the given figure, M is the centre of the circle. Chords AB and CD are perpendicular to each other. If ∠MAD = x and ∠BAC = y.
(i) express ∠AMD in terms of x.
(ii) express ∠ABD in terms of y.
(iii) prove that : x = y

[Diagram: This diagram shows a circle with centre M. Two chords AB and CD cross each other at right angles at point L inside the circle. Points A, B, C, and D are on the circle. Angle MAD is marked as x, and angle BAC is marked as y.]

Answer:
In the figure, M is the centre of the circle.
Chords AB and CD are perpendicular to each other at L.
∠MAD = x and ∠BAC = y

(i) In ΔAMD,
MA = MD (radii of the same circle)
∴ ∠MAD = ∠MDA = x
But in ΔAMD,
∠MAD + ∠MDA + ∠AMD = 180°
⇒ x + x + ∠AMD = 180°
⇒ 2x + ∠AMD = 180°
⇒ ∠AMD = 180° - 2x

(ii) ∴ arc AD ∠AMD at the centre and ∠ABD at the remaining
(Angle in the same segment)
(Angle at the centre is double the angle at the circumference subtended by the same chord)
⇒ ∠AMD = 2∠ABD
⇒ ∠ABD = \( \frac{1}{2} \) ∠AMD
⇒ ∠ABD = \( \frac{1}{2} \)(180° - 2x)
⇒ ∠ABD = 90° - x

AB ⊥ CD, ∠ALC = 90°
In ΔALC,
∴ ∠LAC + ∠LCA = 90°
⇒ ∠BAC + ∠DAC = 90°
⇒ y + ∠DAC = 90°
∴ ∠DAC = 90° - y
We have, ∠DAC = ∠ABD [angles in the same segment]
∴ ∠ABD = 90° - y

(iii) We have, ∠ABD = 90° - y and ∠ABD = 90° - x [proved]
∴ 90° - x = 90° - y
⇒ x = y

📝 Teacher's Note: Show students that equal radii make equal angles. When two chords cross at right angles, special angle relationships are formed. Draw the diagram carefully on the board.

🎯 Exam Tip: Always state that MA = MD because they are radii. Write each step clearly. In part (iii), show both expressions for ∠ABD are equal to prove x = y.

Question 61 (old). In a circle, with centre O, a cyclic quadrilateral ABCD is drawn with AB as a diameter of the circle and CD equal to radius of the circle. If AD and BC produced meet at point P; show that ∠APB = 60°.

[Diagram: This diagram shows a circle with centre O. ABCD is a cyclic quadrilateral with AB as the diameter. CD equals the radius length. Lines AD and BC are extended to meet at point P outside the circle.]

Answer:
Join OD and OC.
In ΔOCD, OD = OC = CD
∴ ΔOCD is an equilateral triangle
∴ ∠ODC = 60°
Also, in cyclic quadrilateral ABCD,
∠ADC + ∠ABC = 180° (Pair of opposite angles in a cyclic quadrilateral are supplementary)
⇒ ∠ODA + 60° + ∠ABP = 180°
⇒ ∠OAD + ∠ABP = 120° (∵ OA = OD)
⇒ ∠PAB + ∠ABP = 120°
By angle sum property of ΔPAB,
∴ ∠APB = 180° - ∠PAB - ∠ABP = 180° - 120° = 60°

📝 Teacher's Note: Emphasize that when all three sides of a triangle are equal, it is equilateral with each angle = 60°. Draw the triangle OCD separately to show this clearly.

🎯 Exam Tip: State clearly that OD = OC = CD = radius. Write "equilateral triangle" and "60° each angle". Use the cyclic quadrilateral property about opposite angles being supplementary.

Exercise 17 B

Question 1. In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal. Prove it.

[Diagram: This diagram shows a cyclic trapezium ABCD inscribed in a circle, where AB is parallel to DC, and AC and BD are the diagonals.]

Answer:
A cyclic trapezium ABCD in which AB || DC and AC and BD are joined.
To prove –
(i) AD = BC
(ii) AC = BD
Proof:
∵ Chord AD subtends ∠ABD and chord BC subtends ∠BDC at the circumference of the circle.
But ∠ABD = ∠BDC [proved]
Chord AD = Chord BC
⇒ AD = BC
Now in ΔADC and ΔBCD
DC = DC [common]
∠CAD = ∠CBD [angles in the same segment]
and AD = BC [proved]
By Side – Angle – Side criterion of congruence, we have
∴ ΔADC ≅ ΔBCD [SAS axiom]
The corresponding parts of the congruent triangles are congruent.
∴ AC = BD [c.p.c.t]

📝 Teacher's Note: Explain that equal chords subtend equal angles. Show students how to identify angles in the same segment. Draw both triangles separately to show congruence clearly.

🎯 Exam Tip: Write "angles in the same segment" for ∠CAD = ∠CBD. State SAS congruence clearly. Always write "c.p.c.t" after proving congruence to get the required result.

Question 2. In the following figure, AD is the diameter of the circle with centre 0. chords AB, BC and CD are equal. If ∠DEF = 110°, calculate:
(i) ∠AFE,
(ii) ∠FAB.

[Diagram: This diagram shows a circle with centre O and diameter AD. Points F, E, B, C are on the circle. Chords AB, BC, and CD are marked as equal. Angle DEF is marked as 110°.]

Answer:
Join AE, OB and OC.
(i) ∵ AOD is the diameter,
∴ ∠AED = 90° [Angle in a semi-circle]
But ∠DEF = 110° [given]
∴ ∠AEF = ∠DEF - ∠AED
= 110° - 90° = 20°
(ii) ∵ Chord AB = Chord BC = Chord CD [given]
∴ ∠AOB = ∠BOC = ∠COD (Equal chords subtends equal angles at the centre)
But ∠AOB + ∠BOC + ∠COD = 180° [AOD is a straight line]
∴ ∠AOB = ∠BOC = ∠COD = 60°
In ΔOAB, OA = OB
∴ ∠OAB = ∠OBA [Radii of the same circle]
But ∠OAB + ∠OBA = 180° - ∠AOB
= 180° - 60°
= 120°

📝 Teacher's Note: Show that when chords are equal, they make equal angles at the centre. Remind students that angle in semicircle is always 90°. Use the fact that angles in isosceles triangle are equal.

🎯 Exam Tip: State clearly "angle in semicircle = 90°" and "equal chords subtend equal angles at centre". Show that three equal angles on a straight line each measure 60°. Write all steps for finding angles in isosceles triangle.

Question 3. If two sides of a cycle-quadrilateral are parallel; prove that:
(i) its other two side are equal.
(ii) its diagonals are equal.
Answer:

[Diagram: This shows a cyclic quadrilateral ABCD inscribed in a circle with center O. AB is parallel to DC, and AC and BD are the diagonals.]

Given:
ABCD is a cyclic quadrilateral in which AB || DC. AC and BD are its diagonals.

To prove:
(i) AD = BC
(ii) AC = BD

Proof:
(i) AB || DC ⇒ ∠DCA = ∠CAB [alternate angles]
Now, chord AD subtends ∠DCA and chord BC subtends ∠CAB at the circumference of the circle.
∴ ∠DCA = ∠CAB [proved]
∴ Chord AD = Chord BC or AD = BC

(ii) Now in ΔABC and ΔADB,
AB = AB [common]
∠ACB = ∠ADB [Angles in the same segment]
BC = AD [proved]
By Side – Angle – Side criterion of congruence, we have
ΔACB ≅ ΔADB [SAS postulate]
The corresponding parts of the congruent triangles are congruent.
∴ AC = BD [c.p.c.t.]

In simple words: When two sides of a four-sided shape inside a circle are parallel, the other two sides become equal. Also, the two diagonal lines become equal length.

📝 Teacher's Note: Draw a cyclic quadrilateral on the board. Show how parallel sides create equal alternate angles. This makes opposite sides equal and diagonals equal.

🎯 Exam Tip: Always write "alternate angles" when you see parallel lines. Then use "angles in same segment are equal" for cyclic quadrilaterals. These are key phrases for full marks.

Question 4. The given figure show a circle with centre O. also, PQ = QR = RS and ∠PTS = 75°. Calculate:
(i) ∠POS,
(ii) ∠ QOR,
(iii) ∠PQR.
Answer:

[Diagram: This shows a circle with center O. Points P, Q, R, S, T are on the circle. PQ, QR, and RS are equal chords.]

Join OP, OQ, OR and OS.

∵ PQ = QR = RS,
∠POQ = ∠QOR = ∠ROS [Equal chords subtends equal angles at the centre]
Arc PQRS subtends ∠POS at the center and ∠PTS at the remaining part of the circle.
∴ ∠POS = 2∠PTS = 2 × 75° = 150°
⇒ ∠POQ + ∠QOR + ∠ROS = 150°
⇒ ∠POQ = ∠QOR = ∠ROS = \(\frac{150°}{3}\) = 50°

In ΔOPQ, OP = OQ [radii of the same circle]
∴ ∠OPQ = ∠OQP
But ∠OPQ + ∠OQP + ∠POQ = 180°
∴ ∠OPQ + ∠OQP = 50° = 180°
⇒ ∠OPQ + ∠OQP = 180° − 50°
⇒ ∠OPQ + ∠OPQ = 130°
⇒ 2∠OPQ = 130°
⇒ ∠OPQ = ∠OQP = \(\frac{130°}{2}\) = 65°

Similarly we can prove that
In ΔOQR, ∠OQR = ∠ORQ = 65°
and in ΔORS, ∠ORS = ∠OSR = 65°

(i) Now ∠POS = 150°
(ii) ∠ QOR = 50° and
(iii) ∠PQR = ∠PQO + ∠OQR = 65° + 65° = 130°

In simple words: Equal chords make equal angles at the center. We use this fact to find all the angles step by step.

📝 Teacher's Note: Show students that equal chords always make equal central angles. Use a compass to draw equal chords and measure the angles to prove this.

🎯 Exam Tip: Write "equal chords subtend equal angles at center" clearly. Show all steps when finding isosceles triangle angles. Don't skip the working.

Question 5. In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of:
(i) ∠ AOB,
(ii) ∠ ACB,
(iii) ∠ABC.
Answer:

[Diagram: This shows a circle with center O. AB is a side of a regular hexagon and AC is a side of a regular octagon, both inscribed in the same circle.]

(i) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.
∴ ∠ACB = \(\frac{1}{2}\) ∠AOB

Since AB is the side of a regular hexagon,
∠AOB = 60°

(ii) ∠AOB = 60° ⇒ ∠ACB = \(\frac{1}{2}\) × 60° = 30°

(iii) Since AC is the side of a regular octagon,
∠AOC = \(\frac{360}{8}\) = 45°

Again, Arc AC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle.
⇒ ∠ABC = \(\frac{1}{2}\) ∠AOC
⇒ ∠ABC = \(\frac{45°}{2}\) = 22.5°

In simple words: For regular polygons, we divide 360° by the number of sides to get the central angle. The angle in the circle is half of the central angle.

📝 Teacher's Note: Make students remember that hexagon has central angle 60° and octagon has 45°. These come up often in exams.

🎯 Exam Tip: Always write the formula: central angle = 360°/number of sides. Then use "angle in alternate segment = half of central angle" for full marks.

Question 6. In a regular pentagon ABCDE, inscribed in a circle; find ratio between angle EDA and angel ADC.
Answer:

[Diagram: This shows a regular pentagon ABCDE inscribed in a circle with center O.]

Arc AE subtends ∠AOE at the centre and ∠ADE at the remaining part of the circle.
∴ ∠ADE = \(\frac{1}{2}\) ∠AOE
= \(\frac{1}{2}\) × 72°
= 36° [central angle of a regular pentagon at O]

∠ADC = ∠ADB + ∠BDC
= 36° + 36° + 72°
∴ ∠ADE: ∠ADC = 36°: 72° = 1: 2

In simple words: In a regular pentagon, each central angle is 72°. We use this to find the angles inside the pentagon and then compare them.

📝 Teacher's Note: Draw a pentagon and show students how to find central angles. Pentagon has 5 sides, so each central angle is 360°/5 = 72°.

🎯 Exam Tip: For pentagon, remember central angle = 72°. Write the ratio in simplest form (1:2, not 36:72) for full marks.

Question 7. In the given figure. AB = BC = CD and ∠ABC = 132°, calculate:
(i) ∠AEB,
(ii) ∠ AED,
(iii) ∠COD.
Answer:

[Diagram: This shows a circle with center O. Points A, B, C, D, E are on the circle. AB = BC = CD are equal chords, and angle ABC = 132°.]

Since AB = BC = CD (given), these are equal chords.
∴ ∠AOB = ∠BOC = ∠COD [equal chords subtend equal angles at center]

In cyclic quadrilateral ABCE,
∠ABC + ∠AEC = 180° [opposite angles of cyclic quadrilateral]
∴ ∠AEC = 180° − 132° = 48°

Now, ∠AEB is part of ∠AEC
Since AB = BC, arc AB = arc BC
∴ ∠AEB = ∠BEC = \(\frac{48°}{2}\) = 24°

Similarly, since BC = CD, arc BC = arc CD
∴ ∠BEC = ∠CED = 24°

∴ ∠AED = ∠AEB + ∠BEC + ∠CED = 24° + 24° + 24° = 72°

For ∠COD:
Since equal chords subtend equal central angles,
and there are 3 equal chords AB, BC, CD
Let each central angle = x
Then 3x represents the arc ACD
The arc ACD subtends ∠AED = 72° at circumference
∴ Central angle for arc ACD = 2 × 72° = 144°
So 3x = 144° ⇒ x = 48°
∴ ∠COD = 48°

In simple words: Equal chords make equal central angles. We use the fact that opposite angles in a cyclic quadrilateral add up to 180°.

📝 Teacher's Note: Show students that equal chords create equal arcs. Use this to split angles at the circumference equally.

🎯 Exam Tip: Always state "equal chords subtend equal central angles" and "opposite angles of cyclic quadrilateral sum to 180°". These are key theorem statements.

Question 8. In the figure, O is the centre of the circle and the length of arc AB is twice the length of arc BC. If angle AOB = 108°, find: (i) ∠CAB, (ii) ∠ADB.

[Diagram: A circle with center O showing points A, B, C, D on the circumference with various lines connecting them]

Answer:
Given: O is the centre of circle with AB = BC = CD
Also, ∠ABC = 132°

(i) In cyclic quadrilateral ABCE
∠ABC + ∠AEC = 180° [sum of opposite angles]
⟹ 132° + ∠AEC = 180°
⟹ ∠AEC = 180° - 132°
⟹ ∠AEC = 48°
Since AB = BC, ∠AEB = ∠BEC [equal chords subtends equal angles]
∴ ∠AEB = \( \frac{1}{2} \)∠AEC
= \( \frac{1}{2} \) × 48°
= 24°

(ii) Similarly, AB = BC = CD
∠AEB = ∠BEC = ∠CED = 24°
∠AED = ∠AEB + ∠BEC + ∠CED
= 24° + 24° + 24° = 72°

(iii) Arc CD subtends ∠COD at the centre and ∠CED at the remaining part of the circle.
∴ ∠COD = 2∠CED
= 2 × 24°
= 48°

Now, Arc AB = 2 arc BC and ∠AOB = 180°
∴ ∠BOC = \( \frac{1}{2} \)∠AOB
= \( \frac{1}{2} \) × 108°
= 54°

Now, Arc BC subtends ∠BOC at the centre and ∠CAB at the remaining part of the circle.
∴ ∠CAB = \( \frac{1}{2} \)∠BOC
= \( \frac{1}{2} \) × 54°
= 27°

(ii) Again, Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.
∴ ∠ACB = \( \frac{1}{2} \)∠AOB
= \( \frac{1}{2} \) × 108°
= 54°

In cyclic quadrilateral ADBC
∠ADB + ∠ACB = 180° [sum of opposite angles]
⟹ ∠ADB + 54° = 180°
⟹ ∠ADB = 180° - 54°
⟹ ∠ADB = 126°

In simple words: We use two main rules here. First, when an arc cuts the circle, the angle at the centre is twice the angle at any point on the circle. Second, opposite angles in a cyclic quadrilateral (a four-sided shape inside a circle) always add up to 180°.

📝 Teacher's Note: Draw a circle on the board and show students how the same arc makes different angles at different points. The centre angle is always double the circumference angle. This is a key circle property students must remember.

🎯 Exam Tip: Always write "sum of opposite angles in cyclic quadrilateral = 180°" and "angle at centre = 2 × angle at circumference". These exact phrases get you marks. Draw the diagram clearly and mark all given angles.

Question 9. The figure shows a circle with centre O. AB is the side of regular pentagon and AC is the side of regular hexagon. Find the angles of triangle ABC.

[Diagram: A circle with center O showing triangle ABC inscribed in the circle]

Answer:
Join OA, OB and OC
Since AB is the side of a regular pentagon,
∠AOB = \( \frac{360°}{5} \) = 72°

Again AC is the side of a regular hexagon,
∠AOC = \( \frac{360°}{6} \) = 60°

But ∠AOB + ∠AOC + ∠BOC = 360° [angles at a point]
⟹ 72° + 60° + ∠BOC = 360°
⟹ 132° + ∠BOC = 360°
⟹ ∠BOC = 360° - 132°
⟹ ∠BOC = 228°

Now, Arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.
⟹ ∠BAC = \( \frac{1}{2} \)∠BOC
⟹ ∠BAC = \( \frac{1}{2} \) × 228° = 114°

Similarly we can prove that
⟹ ∠ABC = \( \frac{1}{2} \)∠AOC
⟹ ∠ABC = \( \frac{1}{2} \) × 60° = 30°

and
⟹ ∠ACB = \( \frac{1}{2} \)∠AOB
⟹ ∠ACB = \( \frac{1}{2} \) × 72° = 36°

Thus, angles of the triangle are, 114°, 30° and 36°

In simple words: In a regular pentagon, each central angle is 360°÷5 = 72°. In a regular hexagon, each central angle is 360°÷6 = 60°. Then we use the rule that angle at circumference is half the angle at centre.

📝 Teacher's Note: Explain that a regular pentagon has 5 equal sides, so the circle is divided into 5 equal parts. Each part has angle 360°÷5. Same logic for hexagon with 6 sides. This makes the concept clear.

🎯 Exam Tip: For regular polygons, always write "Central angle = 360°/n" where n is number of sides. Then use "Angle at circumference = Half of angle at centre". Check that all three angles of triangle add up to 180°.

Question 10. In the given figure, BD is a side of a regular hexagon, DC is a side of a regular pentagon and AD is diameter. Calculate: (i) ∠ADC (ii) ∠BAD, (iii) ∠ABC (iv) ∠AEC.

[Diagram: A circle showing points A, B, C, D, E with AD as diameter and various connecting lines]

Answer:
Join BC, BO, CO and EO
Since BD is the side of a regular hexagon,
∠BOD = \( \frac{360}{6} \) = 60°

Since DC is the side of a regular pentagon,
∠COD = \( \frac{360}{5} \) = 72°

In △BOD, ∠BOD = 60° and ob = od
∴ ∠OBD = ∠ODB = 60°

(i) In △OCD, ∠COD = 72° and OC = OD
∴ ∠ODC = \( \frac{1}{2} \)(180° - 72°)
= \( \frac{1}{2} \) × 108°
= 54°
Or, ∠ADC = 54°

(ii) ∠BDO = 60° or ∠BDA = 60°

(iii) Arc AC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle.
∴ ∠ABC = \( \frac{1}{2} \)∠AOC
= \( \frac{1}{2} \)[∠AOD - ∠COD]
= \( \frac{1}{2} \) × (180° - 72°)
= \( \frac{1}{2} \) × 108°
= 54°

(iv) In cyclic quadrilateral AECD
∠AEC + ∠ADC = 180° [sum of opposite angles]
⟹ ∠AEC + 54° = 180°
⟹ ∠AEC = 180° - 54°
⟹ ∠AEC = 126°

In simple words: We know AD is a diameter, so ∠AOD = 180°. We find angles using regular polygon rules (360°÷6 for hexagon, 360°÷5 for pentagon). Then we use circle angle properties to find the required angles.

📝 Teacher's Note: Remind students that when one side is a diameter, the angle at centre is 180°. Also, in an isosceles triangle (like △OCD where OC = OD), the base angles are equal. This helps find many angles quickly.

🎯 Exam Tip: Write down all given information first: "BD = side of regular hexagon", "DC = side of regular pentagon", "AD = diameter". Then find central angles using 360°/n formula. Show each step clearly for full marks.

Exercise 17 C

Question 1. In the given circle with diameter AB, find the value of x.
Answer:
Step 1: Use angle in same segment property.
\( \angle ABD = \angle ACD = 30° \) (Angle in the same segment)

Step 2: Apply angle sum in triangle ADB.
In \( \triangle ADB \):
\( \angle BAD + \angle ADB + \angle DBA = 180° \) (Angles of a triangle)

Step 3: Use angle in semicircle.
\( \angle ADB = 90° \) (Angle in a semi-circle)

Step 4: Substitute and solve.
\( x + 90° + 30° = 180° \)
\( \implies x + 120° = 180° \)
\( \implies x = 180° - 120° = 60° \)

Therefore, x = 60°
In simple words: When a line cuts a circle into two parts, any angle made on the curved line is 90°. We used this rule and the fact that all angles in a triangle add up to 180°.

📝 Teacher's Note: Show students that any angle made on a semicircle is always 90°. Draw many examples on the board. This is a very important circle property.

🎯 Exam Tip: Always write "angle in semicircle = 90°" and "angles of triangle = 180°". These are the key statements examiners look for.

Question 2. In the given figure, O is the centre of the circle with radius 5 cm, OP and OQ are perpendiculars to AB and CD respectively. AB = 8cm and CD = 6cm. Determine the length of PQ.
Answer:
Given:
Radius of circle = 5 cm
OP ⊥ AB and OQ ⊥ CD
AB = 8 cm and CD = 6 cm

Step 1: Find positions.
Join OA and OC, then OA = OC = 5 cm
Since OP ⊥ AB, P is the midpoint of AB
Similarly Q is the midpoint of CD

Step 2: Calculate OP using right triangle OAP.
In right \( \triangle OAP \):
\( OA^2 = OP^2 + AP^2 \) [Pythagoras Theorem]
\( (5)^2 = OP^2 + (4)^2 \) [∵ AP = PB = \( \frac{1}{2} \times 8 = 4 \) cm]
\( 25 = OP^2 + 16 \)
\( OP^2 = 25 - 16 = 9 \)
\( OP = 3 \) cm

Step 3: Calculate OQ using right triangle OCQ.
In right \( \triangle OCQ \):
\( OC^2 = OQ^2 + CQ^2 \) [Pythagoras Theorem]
\( (5)^2 = OQ^2 + (3)^2 \)
\( 25 = OQ^2 + 9 \)
\( OQ^2 = 25 - 9 = 16 \)
\( OQ = 4 \) cm

Step 4: Find PQ.
Hence, PQ = OP + OQ = 3 + 4 = 7 cm

Therefore, PQ = 7 cm
In simple words: We found how far the center is from each chord. Then we added these distances to get PQ. The perpendicular from center to chord always goes to the middle of the chord.

📝 Teacher's Note: Draw a circle and show that perpendicular from center to any chord cuts the chord exactly in half. This is a key property students must remember.

🎯 Exam Tip: Always write "perpendicular from center bisects the chord" and use Pythagoras theorem. Draw a clear diagram showing right angles.

Question 3. In the given figure, ABC is a triangle in which ∠BAC = 30°. Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose centre is O.
Answer:
Given: In triangle ABC, ∠A = 30°
To prove: BC is the radius of circumcircle of triangle ABC whose centre is O
Construction: Join OB and OC

Proof:
Step 1: Find central angle.
\( \angle BOC = 2 \angle BAC = 2 \times 30° = 60° \)
(Central angle is twice the inscribed angle)

Step 2: Show triangle OBC is equilateral.
Now in \( \triangle OBC \):
OB = OC [Radii of the same circle]
\( \angle OBC = \angle OCB \) [Base angles of isosceles triangle]

Step 3: Calculate base angles.
In \( \triangle BOC \):
\( \angle OBC + \angle OCB + \angle BOC = 180° \) [Angles of a triangle]
\( \angle OBC + \angle OBC + 60° = 180° \)
\( 2 \angle OBC = 180° - 60° = 120° \)
\( \angle OBC = \frac{120°}{2} = 60° \)

Step 4: Conclude equilateral triangle.
\( \angle OBC = \angle OCB = \angle BOC = 60° \)
\( \implies \triangle BOC \) is an equilateral triangle
\( \implies BC = OB = OC \)

But OB and OC are radii of the circumcircle.
∴ BC is also the radius of the circumcircle.

In simple words: When the angle at the top of a triangle is 30°, the bottom side becomes equal to the radius of the circle around the triangle. This happens because all angles in the small triangle at center become 60°.

📝 Teacher's Note: Explain that central angle is always double the inscribed angle. Use a protractor to show students that 60° + 60° + 60° = 180° makes an equilateral triangle.

🎯 Exam Tip: Write "central angle = 2 × inscribed angle" and prove the triangle is equilateral by showing all angles are 60°. This gets full marks.

Question 4. Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Answer:
Given: In triangle ABC, AB = AC and a circle with AB as diameter is drawn which intersects side BC at D
To prove: D is the midpoint of BC
Construction: Join AD

Proof:
Step 1: Use angle in semicircle.
\( \angle 1 = 90° \) [Angle in a semicircle]

Step 2: Use linear pair.
\( \angle 1 + \angle 2 = 180° \) [Linear pair]
\( 90° + \angle 2 = 180° \)
\( \angle 2 = 90° \)

Step 3: Apply RHS congruency.
Now in right triangles ABD and ACD:
Hyp. AB = Hyp. AC [Given]
Side AD = AD [Common]
∴ By RHS criterion of congruence:
\( \triangle ABD \cong \triangle ACD \)

Step 4: Use CPCT.
The corresponding parts of congruent triangles are congruent
∴ BD = DC [CPCT]
Hence D is the midpoint of BC.

In simple words: When we draw a circle using one equal side as width, it cuts the base exactly in half. This happens because the angle made is 90°, creating two identical right triangles.

📝 Teacher's Note: Cut a paper triangle with equal sides. Fold along the height - it shows the base gets bisected. This visual helps students understand the concept.

🎯 Exam Tip: Write "angle in semicircle = 90°" and use RHS congruency. Then write "CPCT" to prove BD = DC. These are the key steps for marks.

Question 5. The given figure shows two circles with centres A and B and radii 5 cm and 3 cm respectively, touching each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.
Answer:

[Diagram: Two circles - a bigger circle with center A and radius 5 cm, and a smaller circle with center B and radius 3 cm inside it, touching internally. A line perpendicular to AB passes through both circles at points P and Q.]

Given:
Circle with center A has radius 5 cm
Circle with center B has radius 3 cm
The circles touch each other internally
Perpendicular bisector of AB meets bigger circle at P and Q

Step 1: Find distance AB.
Since circles touch internally:
AB = 5 - 3 = 2 cm

Step 2: Find midpoint M of AB.
Let M be the midpoint of AB
AM = MB = \( \frac{AB}{2} = \frac{2}{2} = 1 \) cm

Step 3: Use Pythagoras in triangle APM.
In right triangle APM:
\( AP^2 = AM^2 + PM^2 \)
\( 5^2 = 1^2 + PM^2 \)
\( 25 = 1 + PM^2 \)
\( PM^2 = 24 \)
\( PM = 2\sqrt{6} \) cm

Step 4: Find PQ.
Since M is midpoint of PQ:
PQ = 2 × PM = \( 2 × 2\sqrt{6} = 4\sqrt{6} \) cm

Therefore, PQ = \( 4\sqrt{6} \) cm
In simple words: The two circles touch inside each other. The line that cuts AB in half also passes through the big circle. We used the triangle rule to find how long this line is inside the big circle.

📝 Teacher's Note: Show students two rings where small ring fits inside big ring. The distance between centers is the difference of their sizes when they touch inside.

🎯 Exam Tip: For internal touching circles, write "distance between centers = difference of radii". Use Pythagoras theorem clearly and don't forget to double PM to get PQ.

Question 4. In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.
Answer:
Step 1: Join OE.

Step 2: Since arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.
\( \angle EOC = 2 \angle EBC = 2 \times 65° = 130° \)

Step 3: In triangle OEC, OE = OC [Radii of the same circle]
\( \implies \angle OEC = \angle OCE \)

Step 4: In triangle EOC,
\( \angle OEC + \angle OCE + \angle EOC = 180° \) [Angles of a triangle]
\( \implies \angle OCE + \angle OCE + \angle EOC = 180° \)
\( \implies 2 \angle OCE + 130° = 180° \)
\( \implies 2 \angle OCE = 180° - 130° \)
\( \implies 2 \angle OCE = 50° \)
\( \implies \angle OCE = \frac{50°}{2} = 25° \)

Step 5: Since AC || ED [given]
\( \implies \angle DEC = \angle OCE \) [Alternate angles]
\( \implies \angle DEC = 25° \)

In simple words: When a chord is parallel to a diameter, we can use angle properties of circles and parallel lines to find the required angle. The central angle is twice the angle at the circumference.

[Diagram: A circle with center O, diameter AC horizontal, chord ED parallel to AC, point B on circle with angle CBE marked as 65°]

📝 Teacher's Note: Draw the diagram clearly and show students that when we join center to points on the circle, we get radii which are equal. This makes isosceles triangles that help in solving.

🎯 Exam Tip: Always mention "alternate angles" when using parallel lines. Write the central angle theorem clearly: "Central angle = 2 × angle at circumference".

Question 5. The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
Answer:
Given: ABCD is a cyclic quadrilateral and PQRS is a quadrilateral formed by the angle bisectors of ∠A, ∠B, ∠C and ∠D.

To prove: PQRS is a cyclic quadrilateral.

Proof:
In triangle APD,
\( \angle PAD + \angle ADP + \angle APD = 180° \) ...(1)

Similarly, in triangle BQC,
\( \angle QBC + \angle BCQ + \angle BQC = 180° \) ...(2)

Adding (1) and (2), we get
\( \angle PAD + \angle ADP + \angle APD + \angle QBC + \angle BCQ + \angle BQC = 180° + 180° \)
\( \implies \angle PAD + \angle ADP + \angle QBC + \angle BCQ + \angle APD + \angle BQC = 360° \) ....(3)

But \( \angle PAD + \angle ADP + \angle QBC + \angle BCQ = \frac{1}{2}[\angle A + \angle B + \angle C + \angle D] \)

\( = \frac{1}{2} \times 360° = 180° \)

\( \implies \angle APD + \angle BQC = 360° - 180° = 180° \) [from (3)]

But these are the sum of opposite angles of quadrilateral PRQS.
\( \implies \) Quad. PRQS is a cyclic quadrilateral.

In simple words: When we draw angle bisectors inside a cyclic quadrilateral, they form a new quadrilateral. The opposite angles of this new quadrilateral add up to 180°, which means it is also cyclic.

[Diagram: A cyclic quadrilateral ABCD with angle bisectors forming inner quadrilateral PQRS]

📝 Teacher's Note: Start by reminding students that in a cyclic quadrilateral, opposite angles add to 180°. Then show how angle bisectors create half angles that maintain this property.

🎯 Exam Tip: Write "opposite angles are supplementary" clearly. Show all steps of adding triangles and substituting values. This proves the cyclic property.

Question 6. In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:
(i) ∠BDC
(ii) ∠BEC
(iii) ∠BAC
Answer:
(i) Given that BD is a diameter of the circle.
The angle in a semicircle is a right angle.
\( \implies \angle BCD = 90° \)
Also given that ∠DBC = 58°
In triangle BDC,
\( \angle DBC + \angle BCD + \angle BDC = 180° \)
\( \implies 58° + 90° + \angle BDC = 180° \)
\( \implies 148° + \angle BDC = 180° \)
\( \implies \angle BDC = 180° - 148° \)
\( \implies \angle BDC = 32° \)

(ii) We know that the opposite angles of a cyclic quadrilateral are supplementary.
Thus, in cyclic quadrilateral BECD,
\( \angle BEC + \angle BDC = 180° \)
\( \implies \angle BEC + 32° = 180° \)
\( \implies \angle BEC = 180° - 32° \)
\( \implies \angle BEC = 148° \)

(iii) In cyclic quadrilateral ABEC,
\( \angle BAC + \angle BEC = 180° \)
\( \implies \angle BAC + 148° = 180° \)
\( \implies \angle BAC = 180° - 148° \)
\( \implies \angle BAC = 32° \)

In simple words: When BD is a diameter, angle BCD becomes 90°. We use this to find other angles using triangle properties and cyclic quadrilateral properties (opposite angles add to 180°).

[Diagram: A circle with diameter BD, points A, C, E on the circle, angle DBC marked as 58°]

📝 Teacher's Note: Always start with "angle in semicircle = 90°" when diameter is given. Then use triangle angle sum and cyclic quadrilateral properties step by step.

🎯 Exam Tip: Write "angle in semicircle = 90°" first. Then mention "opposite angles of cyclic quadrilateral are supplementary" for each part. Show all calculation steps clearly.

Question 7. D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Proved that the points B, C, E and D are concyclic.
Answer:
Given: In triangle ABC, AB = AC and D and E are points on AB and AC such that AD = AE. DE is joined.

To prove: B, C, E, D are concyclic.

Proof: In triangle ABC, AB = AC
\( \implies \angle B = \angle C \) [Angles opposite to equal sides]

Similarly, in triangle ADE, AD = AE [given]
\( \implies \angle ADE = \angle AED \) [Angles opposite to equal sides]

In triangle ABC,
\( \frac{AP}{AB} = \frac{AE}{AC} \)
\( \implies DE || BC \)
\( \implies \angle ADE = \angle B \) [corresponding angles]
But \( \angle B = \angle C \) [proved]
\( \implies \text{Ext.} \angle ADE = \text{its interior opposite} \angle C \)
\( \implies \) BCED is a cyclic quadrilateral.
Hence B, C, E and D are concyclic.

In simple words: When we have equal sides in triangles, the angles opposite to them are equal. Using these equal angles and parallel line properties, we can show that four points lie on a circle.

[Diagram: Isosceles triangle ABC with points D on AB and E on AC, where AD = AE, showing the quadrilateral BCED]

📝 Teacher's Note: Emphasize that in isosceles triangles, angles opposite equal sides are equal. Show students how parallel lines create corresponding angles that help prove concyclic points.

🎯 Exam Tip: Start with "angles opposite equal sides are equal" for both triangles ABC and ADE. Then use parallel line properties and write "exterior angle equals interior opposite angle" to prove concyclic.

Question 7 (old). Chords AB and CD of a circle intersect each other at point P such that AP = CP. Show that: AB = CD.
Answer:
Given: Chords AB and CD intersect at point P, and AP = CP.

To prove: AB = CD

Proof: Since AP = CP, point P is equidistant from A and C. In a circle, when two chords intersect, if the segments from the intersection point to one pair of endpoints are equal, then the chords are equal.

By the intersecting chords theorem:
AP × PB = CP × PD

Since AP = CP (given), we have:
AP × PB = AP × PD
\( \implies \) PB = PD

Therefore, AP = CP and BP = DP
\( \implies \) AB = AP + PB = CP + PD = CD

Hence, AB = CD.

In simple words: When two chords cross inside a circle and the distances from crossing point to one pair of ends are equal, then the whole chords are equal in length.

📝 Teacher's Note: Use the intersecting chords theorem: when chords cross, the products of their segments are equal. This leads to equal segments and thus equal chords.

🎯 Exam Tip: Write the intersecting chords theorem clearly: AP × PB = CP × PD. Then substitute the given condition AP = CP to show PB = PD, leading to AB = CD.

Question 7. Two chords AB and CD intersect each other at P inside the circle with centre O and AP = CP. Prove that AB = CD.

[Diagram: Circle with center O showing two chords AB and CD intersecting at point P inside the circle]

Answer:
Given: Two chords AB and CD intersect each other at P inside the circle with centre O and AP = CP
To prove: AB = CD
Proof: Two chords AB and CD intersect each other inside the circle at P.
∴ AP × PB = CP × PD
⇒ \( \frac{AP}{CP} = \frac{PD}{PB} \)
But AP = CP ... (1) [given]
∴ PD = PB or PB = PD ... (2)
Adding (1) and (2)
AP + PB = CP + PD
⇒ AB = CD
In simple words: When two chords cross inside a circle, they create four parts. If two parts are equal (AP = CP), then the other two parts are also equal (PB = PD). So the whole chords become equal.

📝 Teacher's Note: Draw two sticks crossing inside a circle. Show students that if one pair of parts is equal, the other pair must also be equal. This is a special property of intersecting chords.

🎯 Exam Tip: Always write the intersecting chords property first: AP × PB = CP × PD. Then use the given condition to prove what is asked.

Question 8. In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ADC = 92°, ∠FAE = 20°; determine ∠BCD. Given reason in support of your answer.

[Diagram: Cyclic quadrilateral ABCD with AF parallel to CB, DA extended to E, showing angles 92° and 20°]

Answer:
In cyclic quad. ABCD,
AF || CB and DA is produced to E such that ∠ADC = 92° and ∠FAE = 20°
Now we need to find the measure of ∠BCD
In cyclic quad. ABCD,
∠B + ∠D = 180°
⇒ ∠B + 92° = 180°
⇒ ∠B = 180° - 92°
⇒ ∠B = 88°
Since AF || CB, ∠FAB = ∠B = 88°
But, ∠FAE = 20° [given]
Ext. ∠BAE = ∠BAF + ∠FAE
= 88° + 20° = 108°
But, Ext. ∠BAE = ∠BCD
∴ ∠BCD = 108°
In simple words: In a cyclic quadrilateral, opposite angles add up to 180°. We used parallel lines and exterior angle properties to find the required angle.

📝 Teacher's Note: Remind students that in a cyclic quadrilateral, opposite angles are supplementary (add to 180°). Also teach the exterior angle property clearly.

🎯 Exam Tip: Write "In cyclic quadrilateral, opposite angles are supplementary" first. Then use parallel line properties. Show all steps clearly for full marks.

Question 9. If I is the in centre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in points D. If ∠BAC = 66° and ∠ABC = 80°, calculate:
(i) ∠DBC
(ii) ∠IBC
(iii) ∠BIC.

[Diagram: Triangle ABC inscribed in a circle with incenter I and AI extended to meet circle at D]

Answer:
Join DB and DC, IB and IC,
∠BAC = 66°, ∠ABC = 80°, I is the incentre of the △ABC,

(i) Since ∠DBC and ∠DAC are in the same segment,
∠DBC = ∠DAC
But, ∠DAC = \( \frac{1}{2} \)∠BAC = \( \frac{1}{2} \) × 66° = 33°
∴ ∠DBC = 33°

(ii) Since I is the incentre of △ABC, IB bisects ∠ABC
∴ ∠IBC = \( \frac{1}{2} \)∠ABC = \( \frac{1}{2} \) × 80° = 40°

(iii) ∠BAC = 66° and ∠ABC = 80°
In △ABC, ∠ACB = 180° - (∠ABC + ∠BAC)
⇒ ∠ACB = 180° - (80° + 66°)
⇒ ∠ACB = 180° - (156°)
⇒ ∠ACB = 34°
Since IC bisects the ∠C,
∴ ∠ICB = \( \frac{1}{2} \)∠C = \( \frac{1}{2} \) × 34° = 17°
Now in △IBC,
∠IBC + ∠ICB + ∠BIC = 180°
⇒ 40° + 17° + ∠BIC = 180°
⇒ 57° + ∠BIC = 180°
⇒ ∠BIC = 180° - 57°
⇒ ∠BIC = 123°

In simple words: The incenter is where all angle bisectors meet. Angles in the same segment of a circle are equal. We used these properties to find the required angles step by step.

📝 Teacher's Note: Explain that the incenter divides each angle into two equal parts. Show students how angles in the same segment are equal using the circle diagram.

🎯 Exam Tip: Always state "I is the incenter, so it bisects angles" and "angles in same segment are equal." Show each calculation step clearly for all three parts.

Question 10. In the given figure, AB = AD = DC = PB and ∠DBC = x°. Determine, in terms of x:
(i) ∠ABD
(ii) ∠APB
Hence or otherwise, prove that AP is parallel to DB.

[Diagram: Circle with points A, B, C, D on circumference and point P above, with equal lengths marked]

Answer:
Given - In the figure, AB = AD = DC = PB and ∠DBC = x°
Join AC and BD.
To find: the measure of ∠ABD and ∠APB.
Proof: ∠DAC = ∠DBC = X [angles in the same segment]
But ∠DCA = ∠DAC = X [∵ AD = DC]
Also, we have, ∠ABD = ∠DAC [angles in the same segment]
In △ABP, ext. ∠ABC = ∠BAP + ∠APB
But, ∠BAP = ∠APB [∵ AB = BP]
2 × X = ∠APB + ∠APB = 2∠APB
∴ 2∠APB = 2X
⇒ ∠APB = X
∴ ∠APB = ∠DBC = X,
But these are corresponding angles
∴ AP || DB

(i) ∠ABD = x°
(ii) ∠APB = x°

In simple words: When sides are equal in triangles, the opposite angles are also equal. We used this property along with angles in the same segment to find the required angles and prove the lines are parallel.

📝 Teacher's Note: Show students that equal sides create equal angles (isosceles triangle property). Also explain how corresponding angles prove parallel lines.

🎯 Exam Tip: Write "angles in the same segment are equal" and "equal sides give equal angles." Show that corresponding angles are equal to prove parallel lines.

Question 11. In the given figure; ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.

[Diagram: Two intersecting circles with straight lines ABC, AEQ, and CEP marked]

Answer:
Given: ABC, AEQ and CEP are straight lines
To prove: ∠APE and ∠CQE are supplementary
Proof:
Since APEC is a cyclic quadrilateral (all four points lie on a circle),
∠APE + ∠ACE = 180° ... (1) [opposite angles of cyclic quadrilateral]
Since AQEC is also a cyclic quadrilateral,
∠CQE + ∠CAE = 180° ... (2) [opposite angles of cyclic quadrilateral]
But ∠ACE = ∠CAE [same angle]
From (1) and (2):
∠APE + ∠ACE = ∠CQE + ∠CAE
Since ∠ACE = ∠CAE,
∠APE = ∠CQE
But in cyclic quadrilateral APEC:
∠APE + ∠ACE = 180°
Therefore, ∠APE + ∠CQE = 180°
Hence, ∠APE and ∠CQE are supplementary.

In simple words: When four points lie on a circle, opposite angles add up to 180°. We used this property for both sets of four points to show the required angles are supplementary.

📝 Teacher's Note: Draw two intersecting circles clearly and mark the cyclic quadrilaterals. Explain that opposite angles in a cyclic quadrilateral always add to 180°.

🎯 Exam Tip: Identify the cyclic quadrilaterals first. Write "opposite angles of cyclic quadrilateral are supplementary" and apply this property step by step.

Question 12. In the given, AB is the diameter of the circle with centre O.
Answer:

[Diagram: A circle with diameter AB and center O. Points P, Q, E, C are marked on the circle forming a geometric construction with straight lines ABC, AEQ and CEP.]

Given: In the figure, ABC, AEQ and CEP are straight lines
To prove: \( \angle APE + \angle CQE = 180° \)
Construction: Join EB
Proof: In cyclic quad ABEP,
\( \angle APE + \angle ABE = 180° \) ......(1)
Similarly, in cyclic quad BCQE,
\( \angle CQE + \angle CBE = 180° \) ......(2)
Adding (1) and (2),
\( \angle APE + \angle ABE + \angle CQE + \angle CBE = 180° + 180° = 360° \)

\( \implies \angle APE + \angle ABE + \angle CBE = 360° \)
But, \( \angle ABE + \angle CBE = 180° \) [Linear pair]

\( \therefore \angle APE + \angle CQE + 180° = 360° \)

\( \implies \angle APE + \angle CQE = 360° - 180° = 180° \)
Hence \( \angle APE \) AND \( \angle CQE \) are supplementary.

In simple words: We used the property that opposite angles in a cyclic quadrilateral add up to 180°. By making two cyclic quadrilaterals and adding their equations, we proved the angles are supplementary.

📝 Teacher's Note: Draw the circle and mark all points clearly. Show students how we make two separate cyclic quadrilaterals from the same figure. This is a common technique in circle geometry.

🎯 Exam Tip: Always write "cyclic quadrilateral" clearly and state the property that opposite angles sum to 180°. Show the linear pair property when angles are on a straight line.

Question 12. In the given, AB is the diameter of the circle with centre O. If ∠ADC = 32°, find angle BOC.
Answer:

[Diagram: A circle with diameter AB, center O, and points C and D on the circle. Angle ADC is marked as 32°.]

Arc AC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle

\( \therefore \angle AOC = 2 \angle ADC \)

\( \implies \angle AOC = 2 \times 32° = 64° \)
Since ∠AOC and ∠BOC are linear pair, we have
\( \angle AOC + \angle BOC = 180° \)

\( \implies 64° + \angle BOC = 180° \)

\( \implies \angle BOC = 180° \)

\( \implies \angle BOC = 180° - 64° \)

\( \implies \angle BOC = 116° \)

In simple words: The angle at the center is double the angle at any point on the circle. We used this rule and the fact that angles on a straight line add to 180°.

📝 Teacher's Note: Emphasize the center-angle theorem: angle at center = 2 × angle at circumference. Also show that AB being diameter means AOB is a straight line, so AOC + BOC = 180°.

🎯 Exam Tip: State the theorem clearly: "Angle at center is twice the angle at circumference." Then show the linear pair calculation step by step.

Question 13. In a cyclic-quadrilateral PQRS, angle PQR = 135°. Sides SP and RQ produced meet at point A: whereas sides PQ and SR produced meet at point B. If ∠A: ∠B = 2 : 1, find angles A and B.
Answer:

[Diagram: A cyclic quadrilateral PQRS with angle PQR = 135°. Extended sides meet at points A and B outside the circle.]

PQRS is a cyclic quadrilateral in which ∠PQR = 135°
Sides SP and RQ are produced to meet at A and
Sides PQ and SR are produced to meet at B.
∠A : ∠B = 2 : 1
Let ∠A = 2x, then ∠B = x
Now, in cyclic quad PQRS,
Since, ∠PQR = 135°, ∠S = 180° - 135° = 45°
[Since sum of opposite angles of a cyclic quadrilateral are supplementary]
Since, ∠PQR and ∠PQA are linear pair,
∠PQR + ∠PQA = 180°

\( \implies 135° + \angle PQA = 180° \)

\( \implies \angle PQA = 180° - 135° = 45° \)
Now, in △PBS,
∠P = 180° - (45° + x) = 180° - 45° - x = 135° - x ....(1)
Again, in △PQA,
Ext.∠P = ∠PQA + ∠A = 45° + 2x .....(2)
From (1) and (2),
45° + 2x = 135° - x

\( \implies 2x + x = 135° - 45° \)

\( \implies 3x = 90° \)

\( \implies x = 30° \)
Hence, ∠A = 2x = 2 × 30° = 60°
and ∠B = x = 30°

In simple words: We used the property that opposite angles in a cyclic quadrilateral add to 180°. Then we used triangle angle sum and exterior angle properties to find the unknown angles.

📝 Teacher's Note: Draw the figure carefully showing the extended lines meeting outside. Explain that exterior angle of a triangle equals sum of two opposite interior angles. This is key to solving this problem.

🎯 Exam Tip: Write the ratio clearly as equations: ∠A = 2x, ∠B = x. State the cyclic quadrilateral property and exterior angle theorem. Show all algebraic steps clearly.

Question 17 (old). If the following figure, AB is the diameter of a circle with centre O and CD is the chord with length equal radius OA. If AC produced and BD produced meet at point P, show that ∠APB = 60°
Answer:

[Diagram: A circle with diameter AB, center O, and chord CD equal to radius OA. Extended lines AC and BD meet at point P outside the circle.]

Given: In the figure, AB is the diameter of a circle with centre O.
CD is the chord with length equal radius OA.
AC and BD produced meet at point P
To prove: ∠APB = 60°
Construction: Join OC and OD
Proof: We have CD = OC = OD [given]
Therefore, △OCD is an equilateral triangle

\( \therefore \angle OCD = \angle ODC = \angle COD = 60° \)
In △AOC, OA = OC [radii of the same circle]

\( \therefore \angle A = \angle 1 \)
Similarly, in △BOD, OB = OD [radii of the same circle]

\( \therefore \angle B = \angle 2 \)
Now, in cyclic quad ACDB,

since, ∠ACD + ∠B = 180°
[Since sum of opposite angles of a cyclic quadrilateral are supplementary]

\( \implies 60° + \angle 1 + \angle B = 180° \)

\( \implies \angle 1 + \angle B = 180° - 60° \)

\( \implies \angle 1 + \angle B = 120° \)
But, ∠1 = ∠A

\( \therefore \angle A + \angle B = 120° \) ....(1)
Now, in △APB,
∠P + ∠A + ∠B = 180° [Sum of angles of a triangle]

\( \implies \angle P + 120° = 180° \)

\( \implies \angle P = 180° - 120° \) [from (1)]

\( \implies \angle P = 60° \) or ∠APB = 60°

In simple words: Since CD equals the radius, triangle OCD becomes equilateral with all angles 60°. Using cyclic quadrilateral properties and triangle angle sum, we proved ∠APB = 60°.

📝 Teacher's Note: Show students that when a chord equals the radius, it creates an equilateral triangle with the center. This is a special case that often appears in geometry problems.

🎯 Exam Tip: State clearly that CD = radius makes triangle OCD equilateral. Write the cyclic quadrilateral property and show the triangle angle sum calculation step by step.

Question 14. In the following figure, ABCD is a cyclic quadrilateral in which AD is parallel to BC. If the bisector of angle A meet BC at point E and the given circle at point F, prove that: (i) EF = FC (ii) BF = DF
Answer:

[Diagram: A cyclic quadrilateral ABCD with AD parallel to BC. The angle bisector of angle A meets BC at E and the circle at F.]

Solution:
(i) To prove EF = FC:
Since AF is the bisector of ∠DAB,
∠DAF = ∠BAF
Since AD || BC,
∠DAF = ∠AFC (alternate angles)
Also, ∠BAF = ∠BFA (angles in the same segment)
Therefore, ∠AFC = ∠BFA
In triangle EFC, ∠EFC = ∠ECF
Therefore, EF = EC
(ii) To prove BF = DF:
Since ABCD is a cyclic quadrilateral and AD || BC,
ABCD is an isosceles trapezium
Therefore, AB = DC
Since AF bisects ∠BAD,
∠BAF = ∠DAF
In triangles ABF and ADF:
AB = DC (proved above)
AF = AF (common)
∠BAF = ∠DAF (AF is angle bisector)
Therefore, triangle ABF ≅ triangle ADF (SAS)
Hence, BF = DF

In simple words: When we have a cyclic quadrilateral with parallel sides, the angle bisector creates equal segments. We used properties of parallel lines and congruent triangles to prove this.

📝 Teacher's Note: Explain that a cyclic quadrilateral with one pair of parallel sides is always an isosceles trapezium. The angle bisector creates symmetry in the figure.

🎯 Exam Tip: For part (i), use alternate angles and angles in same segment. For part (ii), prove the triangles are congruent using SAS. Write all properties clearly.

Question 15. ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point E; whereas sides BC and AD produced meet at point F. If ∠DCF : ∠F : ∠E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.
Answer:
Given: In a circle, ABCD is a cyclic quadrilateral AB and DC are produced to meet at E and BC and AD are produced to meet at F.
∠DCF : ∠F : ∠E = 3 : 5 : 4
Let ∠DCF = 3x, ∠F = 5x, ∠E = 4x
Now, we have to find ∠A, ∠B, ∠C AND ∠D
In cyclic quad. ABCD, BC is produced.
∴ ∠A = ∠DCF = 3x
In △CDF,
Ext.∠CDA = ∠DCF + ∠F = 3x + 5x = 8x
In △BCE,
Ext.∠ABC = ∠BCE + ∠E [∠BCE = ∠DCF, vertically opposite angles]
= ∠DCF + ∠E
= 3x + 4x = 7x
Now, in cyclic quad.ABCD,
since, ∠B + ∠D = 180°
[Since sum of opposite of a cyclic quadrilateral are supplementary]
⇒ 7x + 8x = 180°
⇒ 15x = 180°
⇒ x = 180°/15 = 12°
∴ ∠A = 3x = 3 × 12° = 36°
∠B = 7x = 7 × 12° = 84°
∠C = 180° - ∠A = 180° - 36° = 144°
∠D = 8x = 8 × 12° = 96°
In simple words: We used the property that opposite angles in a cyclic quadrilateral add up to 180°. The angles outside the quadrilateral help us find the angles inside it.

[Diagram: This diagram shows a cyclic quadrilateral ABCD inscribed in a circle, with sides extended to meet at points E and F outside the circle.]

📝 Teacher's Note: Draw the figure carefully first. Mark all the angles clearly. Remember that when sides are extended, they form exterior angles which are equal to opposite interior angles in cyclic quadrilaterals.

🎯 Exam Tip: Always write "sum of opposite angles = 180°" when dealing with cyclic quadrilaterals. This is the key property examiners look for. Show all substitution steps clearly.

Question 16. The following figure shows a circle with PR as its diameter. If PQ = 7 cm and QR = 3RS = 6 cm, Find the perimeter of the cyclic quadrilateral PQRS.
Answer:
In the figure, PQRS is a cyclic quadrilateral in which PR is a diameter
PQ = 7 cm
QR = 3RS = 6 cm
3RS = 6 cm ⇒ RS = 2 cm
Now in △PQR,
∠Q = 90° [Angles in a semi circle]
∴ PR² = PQ² + QR² [Pythagoras Theorem]
= 7² + 6²
= 49 + 36
= 85
Again in right △PSQ, PR² = PS² + RS²
⇒ 85 = PS² + 2²
⇒ PS² = 85 - 4 = 81 = (9)²
∴ PS = 9 cm
Now, perimeter of quad.PQRS = PQ + QR + RS + SP
= (7 + 9 + 2 + 6) cm
= 24
In simple words: Since PR is a diameter, any angle on the circle looking at PR will be 90°. We used Pythagoras theorem to find the missing side, then added all four sides to get the perimeter.

[Diagram: This diagram shows a circle with diameter PR, and points Q and S on the circle forming quadrilateral PQRS.]

📝 Teacher's Note: Remind students that angle in a semicircle is always 90°. This creates right triangles which we can solve using Pythagoras theorem. Draw the diameter clearly in the diagram.

🎯 Exam Tip: Write "angle in semicircle = 90°" clearly. Then use Pythagoras theorem step by step. Finally, add all four sides for perimeter. Show units (cm) in the final answer.

Question 17. In the following figure, AB is the diameter of a circle with centre O. If chord AC = chord AD. Prove that:
(i) arc BC = arc DB
(ii) AB is bisector of ∠CAD.
Further if the length of arc AC is twice the length of arc BC find:
(a) ∠BAC
(b) ∠ABC
Answer:
Given - In a circle with centre O, AB is the diameter and AC and AD are two chords such that AC = AD.
To prove: (i) arc BC = arc DB
(ii) AB is the bisector of ∠CAD
(iii) If arc AC = 2arc BC, then find
(a) ∠BAC (b) ∠ABC
Construction: Join BC and BD
Proof: In right angled △ABC and △ABD
Side AC = AD [given]
Hyp. AB = AB [common]
∴ By Right Angle – Hypotenuse – Side criterion of congruence,
△ABC ≅ △ABD
(i) The corresponding parts of the congruent triangles are congruent.
∴ BC = BD [c.p.c.t]
∴ Arc BC = ArcBD [equal chords have equal arcs]
(ii) ∠BAC = ∠BAD
∴ AB is the bisector of ∠CAD
(iii) If Arc AC = 2 arc BC,
then ∠ABC = 2∠BAC
But ∠ABC + ∠BAC = 90°
⇒ 2∠BAC + ∠BAC = 90°
⇒ 3∠BAC = 90°
⇒ ∠BAC = 90°/3 = 30°
∠ABC = 2∠BAC ⇒ ∠ABC = 2 × 30° = 60°
In simple words: Equal chords make equal arcs. When AB is a diameter, angles ACB and ADB are 90°. Using congruent triangles, we proved AB bisects angle CAD. Then we used the given ratio to find the specific angles.

[Diagram: This diagram shows a circle with diameter AB and points C and D on the circle, with chords AC and AD being equal in length.]

📝 Teacher's Note: Start by showing that angles in semicircle are 90°. Then use RHS congruence rule. Students often forget to mention "corresponding parts of congruent triangles" - emphasize this step.

🎯 Exam Tip: Write "Given", "To Prove", and "Construction" clearly. For congruence, mention the exact rule used (RHS here). For part (iii), use the fact that inscribed angle = half the arc it subtends.

Question 18. In cyclic quadrilateral ABCD; AD = BC, ∠ = 30° and ∠ = 70°; find;
(i) ∠BCD
(ii) ∠BCA
(iii) ∠ABC
(iv) ∠ADC
Answer:
Note: The question appears to have missing angle labels. Based on the context and typical cyclic quadrilateral problems, I will assume the given angles are ∠BAD = 30° and ∠CAD = 70°.

Given: ABCD is a cyclic quadrilateral, AD = BC, ∠BAD = 30°, ∠CAD = 70°

Since AD = BC (equal chords), arcs AD = BC
This means ∠ABD = ∠BCA [angles subtended by equal arcs]

(i) In cyclic quadrilateral ABCD:
∠BCD + ∠BAD = 180° [opposite angles are supplementary]
∠BCD + 30° = 180°
∠BCD = 150°

(ii) Since equal chords subtend equal angles:
∠BCA = ∠ABD
In △ACD: ∠CAD = 70° (given)
Since AD = BC, we get ∠BCA = 70°

(iii) ∠ABC = ∠ABD + ∠DBC
Using properties of cyclic quadrilateral and equal chords:
∠ABC = 180° - ∠ADC = 180° - 110° = 70°

(iv) ∠ADC + ∠ABC = 180° [opposite angles]
∠ADC = 180° - 70° = 110°

In simple words: In a cyclic quadrilateral, opposite angles add up to 180°. Equal chords create equal arcs and equal angles. We use these two main properties to find all the angles.

📝 Teacher's Note: The original question has incomplete angle labels. Always check that given angles are clearly marked. Use the property that equal chords subtend equal angles at the circumference.

🎯 Exam Tip: Write "opposite angles = 180°" and "equal chords subtend equal angles" as your key statements. Always check if your angles add up correctly for the quadrilateral (sum should be 360°).

Question. ABCD is a cyclic quadrilateral and AD = BC
Answer:

[Diagram: This diagram shows a circle with cyclic quadrilateral ABCD inscribed in it, where AD = BC. The diagram shows various angles marked including ∠BAC = 30°, ∠CBD = 70°.]

Given: ABCD is a cyclic quadrilateral and AD = BC
∠BAC = 30°, ∠CBD = 70°
We have
∠DAC = ∠CBD [angles in the same segment]

\( \implies \) ∠DAC = 70° [∵ ∠CBD = 70°]

\( \implies \) ∠BAD = ∠BAC + ∠DAC = 30° + 70° = 100° .....(1)
Since the sum of opposite angles of cyclic quadrilateral is supplementary
∠BAD + ∠BCD = 180°

\( \implies \) 100° + ∠BCD = 180° [from (1)]

\( \implies \) ∠BCD = 180° - 100° = 80°
Since AD = BC, ∠ACD = ∠BDC [Equal chords subtends equal angles]
But ∠ACB = ∠ADB [angles in the same segment]
∴ ∠ACD + ∠ACB = ∠BDC + ∠ADB

\( \implies \) ∠BCD = ∠ADC = 80°
But in ΔBCD,
∠CBD + ∠BCD + ∠BDC = 180° [angles of a triangle]

\( \implies \) 70° + 80° + ∠BDC = 180°

\( \implies \) 150° + ∠BDC = 180°
∴ ∠BDC = 180° - 150° = 30°

\( \implies \) ∠ACD = 30° [∵ ∠ACD = ∠BDC]
∴ ∠BCA = ∠BCD - ∠ACD = 80° - 30° = 50°
Since the sum of opposite angles of cyclic quadrilateral is supplementary,
∠ADC + ∠ABC = 180°

\( \implies \) 80° + ∠ABC = 180°

\( \implies \) ∠ABC = 180° - 80° = 100°
In simple words: In a cyclic quadrilateral, opposite angles add up to 180°. Equal chords make equal angles at the circumference. We use these facts to find all the angles step by step.

📝 Teacher's Note: Draw a circle on the board and show how opposite angles in a cyclic quadrilateral always add to 180°. This is a key property students must remember.

🎯 Exam Tip: Always state "opposite angles of cyclic quadrilateral are supplementary" and "equal chords subtend equal angles." These key phrases get you marks.

Question 19. In the given figure, ∠ACE = 43° and ∠CAF = 62°; find the values of a, b and c.

[Diagram: This diagram shows a circle with points A, B, C, D, E, F marked. There are angles marked as a, b, c, and given angles of 43° and 62°.]

Answer:
Now, ∠ACE = 43° and ∠CAF = 62° [given]
In ΔAEC,
∴ ∠ACE + ∠CAE + ∠AEC = 180°

\( \implies \) 43° + 62° + ∠AEC = 180°

\( \implies \) 105° + ∠AEC = 180°

\( \implies \) ∠AEC = 180° - 105° = 75°
Now, ∠ABD + ∠AED = 180° [Opposite angles of a cyclic quad and ∠AED = ∠AEC]

\( \implies \) a + 75° = 180°

\( \implies \) a = 180° - 75°

\( \implies \) a = 105°
∠EDF = ∠BAE [angles in the alternate segments]
∴ c = 62°
In ΔBAF, a + 62° + b = 180°

\( \implies \) 105° + 62° + b = 180°

\( \implies \) 167° + b = 180°

\( \implies \) b = 180° - 167° = 13°
Hence, a = 105°, b = 13° and c = 62°
In simple words: We use the fact that angles in a triangle add to 180°, and opposite angles in a cyclic quadrilateral add to 180°. We also use the alternate segment theorem.

📝 Teacher's Note: Teach students to identify which angles are in the same segment and which are alternate segment angles. Use different colored chalk to mark these relationships clearly.

🎯 Exam Tip: Always write "angles in triangle = 180°" and "opposite angles in cyclic quad = 180°." Show each step clearly with proper reasoning.

Question 20. In the given figure, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°
Find
(i) ∠CAD
(ii) ∠CBD
(iii) ∠ADC

[Diagram: This diagram shows a circle with cyclic quadrilateral ABCD where AB is parallel to DC. Angles of 25° and 80° are marked.]

Answer:
In the given figure,
ABCD is a cyclic quad in which AB || DC
∴ ABCD is an isosceles trapezium
∴ AD = BC
(i) Join BD and we have,
Ext. ∠BCE = ∠BAD [Exterior angle of a cyclic qud is equal to interior opposite angle]
∴ ∠BAD = 80° [∵ ∠BCE = 80°]
But ∠BAC = 25°
∴ ∠CAD = ∠BAD - ∠BAC
= 80° - 25°
= 55° (ii) ∠CBD = ∠CAD [angle of the same segment]
= 55° (iii) ∠ADC = ∠BCD [angles of the isosceles trapezium]
= 180° - ∠BCE
= 180° - 80°
= 100°
In simple words: When AB is parallel to DC in a cyclic quadrilateral, it becomes an isosceles trapezium. This means AD = BC and some angles are equal. We use the exterior angle property of cyclic quadrilaterals.

📝 Teacher's Note: Draw a trapezium and show students that when it is inscribed in a circle with parallel sides, it becomes isosceles. This helps them see why certain angles are equal.

🎯 Exam Tip: Write "isosceles trapezium" clearly and state "exterior angle = opposite interior angle" for cyclic quadrilaterals. These are key marking points.

Question 21. ABCD is a cyclic quadrilateral of a circle with centre o such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle if AD and BC produced meet at P, show that ∠APB = 60°

[Diagram: This diagram shows a circle with center O, where AB is a diameter, CD is a chord equal to the radius, and extended lines AD and BC meet at point P outside the circle.]

Answer:
Construction — Join OC and OD
Proof — Since chord CD = CO = DO [radii of the circle]
∴ ΔDOC is an equilateral triangle
∴ ∠DOC = ∠ODC = ∠DCO = 60°
Let ∠A = x and ∠B = y
Since OA = OB = OC = OD [radii of the same circle]
∴ ∠ODA = ∠OAD = x and
∠OCB = ∠OBC = y
∴ ∠AOD = 180° - 2x and ∠BOC = 180° - 2y
But AOB is a straight line
∴ ∠AOD + ∠BOC + ∠COD = 180°

\( \implies \) 180° - 2x + 180° - 2y + 60° = 180°

\( \implies \) 2x + 2y = 240°

\( \implies \) x + y = 120°
But ∠A + ∠B + ∠P = 180° [Angles of a triangle]

\( \implies \) 120° + ∠P = 180°

\( \implies \) ∠P = 180° - 120°

\( \implies \) ∠P = 60°
Hence ∠APB = 60°
In simple words: When a chord equals the radius, it forms an equilateral triangle with the center. Using angle properties and the fact that AB is a diameter, we can prove that P makes a 60° angle.

📝 Teacher's Note: Show students how chord = radius creates an equilateral triangle. Use a compass to demonstrate this property. It helps students visualize the 60° angles.

🎯 Exam Tip: State clearly "chord = radius forms equilateral triangle" and "AB is diameter so angles sum to 180°." Show the construction step by step for full marks.

Question 22. In the figure, given alongside, CP bisects angle ACB. Show that DP bisects angle ADB.

[Diagram: This diagram shows a circle with points A, B, C, D on the circumference, with point P inside where CP bisects angle ACB, and we need to prove DP bisects angle ADB.]

Answer:
Given: CP bisects ∠ACB
To Prove: DP bisects ∠ADB
Since CP bisects ∠ACB,
∠ACP = ∠BCP
∠ADP = ∠ACP [angles in the same segment of circle]
∠BDP = ∠BCP [angles in the same segment of circle]
Since ∠ACP = ∠BCP [CP bisects ∠ACB]

\( \implies \) ∠ADP = ∠BDP
Therefore, DP bisects ∠ADB.
In simple words: When a line bisects one angle in a cyclic quadrilateral, the line from the opposite vertex also bisects the angle there. This happens because angles in the same segment are equal.

📝 Teacher's Note: Draw the circle clearly and show students which angles are in the same segment. Use different colors to mark equal angles. This visual helps them understand the concept easily.

🎯 Exam Tip: Always state "angles in the same segment are equal" as your main reason. Write "Given" and "To Prove" clearly at the start. Show the logical steps one by one.

Question 23. In the figure, given below, AD = BC, ∠BAC = 30° and ∠ = 70° find:
(i) ∠BCD
(ii) ∠BCA
(iii) ∠ABC
(iv) ∠ADC

[Diagram: This diagram shows a cyclic quadrilateral ABCD inscribed in a circle with points A, B, C, D on the circle. AC and BD are diagonals intersecting inside the circle.]

Answer: In the figure, ABCD is a cyclic quadrilateral. AC and BD are its diagonals.
∠BAC = 30° and ∠CBD = 70°
Now we have to find the measures of ∠BCD, ∠BCA, ∠ABC and ∠ADB
We have ∠CAD = ∠CBD = 70° [Angles in the same segment]
Similarly, ∠BAC = ∠BDC = 30°
∴ ∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°
(i) Now ∠BCD + ∠BAD = 180° [opposite angles of cyclic quadrilateral]
⇒ ∠BCD + ∠BAD = 180°
⇒ ∠BCD + 100° = 180°
⇒ ∠BCD = 180° - 100°
⇒ ∠BCD = 80°
(ii) Since AD = BC, ABCD is an isosceles trapezium and AB || DC
∠BAC = ∠DCA [alternate angles]
⇒ ∠DCA = 30°
∴ ∠ABD = ∠DAC = 30° [angles in the same segment]
∴ ∠BCA = ∠BCD - ∠DAC = 80° - 30° = 50°
(iii) ∠ABC = ∠ABD + ∠CBD = 30° + 70° = 100°
(iv) ∠ADB = ∠BCA = 50° [angles in the same segment]
In simple words: In a cyclic quadrilateral (four sides inside a circle), opposite angles always add to 180°. Angles that look at the same arc are equal.

📝 Teacher's Note: Draw a circle on the board. Mark four points and show how opposite angles add to 180°. This is the main property of cyclic quadrilaterals.

🎯 Exam Tip: Always write "angles in the same segment" or "opposite angles of cyclic quadrilateral" as your reason. These exact phrases get you marks.

Question 24. In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find:
(i) ∠OBD
(ii) ∠AOB
(iii) ∠BED (2016)

Answer: i. AD is parallel to BC, i.e., OD is parallel to BC and BD is transversal. ii. AD is parallel to BC, i.e., AO is parallel to BC and OB is transversal.
⇒ ∠AOB = ∠OBC [Alternate angles]
⇒ ∠OBC = ∠OBD + ∠DBC
⇒ ∠OBC = 32° + 32°
⇒ ∠OBC = 64°
⇒ ∠AOB = 64° iii. In △OAB,
OA = OB [Radii of the same circle]
⇒ ∠OAB = ∠OBA = x (say)
⇒ ∠OAB + ∠OBA + ∠AOB = 180°
⇒ x + x + 64° = 180°
⇒ 2x = 180° - 64°
⇒ 2x = 116°
⇒ x = 58°
∴ ∠OAB = 58° i.e., ∠DAB = 58°
⇒ ∠DAB = ∠BED = 58° [Angles inscribed in the same arc are equal]
In △OBD,
OD = OB [Radii of the same circle]
⇒ ∠ODB = ∠OBD = 32°
In simple words: When two lines are parallel, alternate angles are equal. In a circle, radii are equal, so triangles with two radii are isosceles (two equal sides make two equal angles).

📝 Teacher's Note: Show students that all radii of a circle are equal. So any triangle with two radii as sides will have two equal angles at the base.

🎯 Exam Tip: Always write "radii of the same circle" when you say OA = OB. Write "alternate angles" when using parallel line properties. Use exact terms.

Question 25. In the figure given, O is the centre of the circle. ∠DAE = 70°. Find giving suitable reasons, the measure of
i. ∠BCD
ii. ∠BOD
iii. ∠OBD

[Diagram: This diagram shows a circle with center O. Points A, B, C, D, E are on the circle. Lines connect various points, and angle DAE is marked as 70°.]

Answer: ∠DAE and ∠DAB are linear pair
So, ∠DAE + ∠DAB = 180°
∴ ∠DAB = 110° Also, ∠BCD + ∠DAB = 180° [Opp. Angles of cyclic quadrilateral BADC]
∴ ∠BCD = 70° ∠BCD = \( \frac{1}{2} \) ∠BOD [angles subtended by an arc on the centre and on the circle]
∴ ∠BOD = 140° In △BOD,
OB = OD [radii of same circle]
So, ∠OBD = ∠ODB [isosceles triangle theorem]
∠OBD + ∠ODB + ∠BOD = 180° [sum of angles of triangle]
2∠OBD = 40°
∠OBD = 20°
In simple words: Linear pair angles add to 180°. In a cyclic quadrilateral, opposite angles add to 180°. The angle at center is double the angle at circumference for the same arc.

📝 Teacher's Note: Teach students that center angle is always double the circumference angle. Draw a pizza slice to show this - the angle at center of slice is bigger than angle at the edge.

🎯 Exam Tip: Write "linear pair", "opposite angles of cyclic quadrilateral", and "angle at center is double angle at circumference". These are key phrases examiners look for.