GSEB Class 9 Maths Solutions Chapter 10 Circles Exercise 10.5

Question 1. In figure, A, B and C are three points on a circle with centre O such that \( \angle BOC = 30^\circ \) and \( \angle AOB = 60^\circ \). If D is a point on the circle other than the arc ABC, find \( \angle ADC \).
Answer: \( \angle ADC = \frac {1}{2}\angle AOC \) The angle formed by an arc at the centre is double the angle formed by it at any point on the remaining part of the circle.
\( \implies \angle ADC = \frac {1}{2}(\angle AOB + \angle BOC) \)
\( \implies \angle ADC = \frac {1}{2} (60^\circ + 30^\circ) \)
\( \implies \angle ADC = \frac {1}{2} (90^\circ) \)
\( \implies \angle ADC = 45^\circ \)

In simple words: The angle at the centre of a circle, made by an arc, is always twice as big as the angle that same arc makes at any point on the rest of the circle. First, we add the two central angles to get the total angle for arc AC. Then, we divide that total by two to find the angle at point D.

Exam Tip: Remember the theorem: "The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle." This is crucial for solving problems like this.

Question 2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Answer: Given that \( OA = OB = AB \). This means that triangle \( OAB \) is an equilateral triangle.
\( \implies \angle AOB = 60^\circ \)
The angle formed by an arc at the centre is double the angle formed by it at any point on the remaining part of the circle.
\( \implies \angle ACB = \frac {1}{2}\angle AOB \)
\( \implies \angle ACB = \frac {1}{2} \times 60^\circ \)
\( \implies \angle ACB = 30^\circ \)
Now, \( ADBC \) is a cyclic quadrilateral.
The sum of either pair of opposite angles of a cyclic quadrilateral is \( 180^\circ \).
\( \implies \angle ADB + \angle ACB = 180^\circ \)
\( \implies \angle ADB + 30^\circ = 180^\circ \)
\( \implies \angle ADB = 180^\circ - 30^\circ \)
\( \implies \angle ADB = 150^\circ \)

In simple words: When a chord is the same length as the radius, it forms an equilateral triangle with the center, so the angle at the centre is \( 60^\circ \). The angle on the major arc is half of this central angle. The angle on the minor arc is found by subtracting the major arc angle from \( 180^\circ \), because opposite angles in a cyclic quadrilateral always add up to \( 180^\circ \).

Exam Tip: Remember that if a chord equals the radius, it forms an equilateral triangle with the center. Also, keep in mind the cyclic quadrilateral property for opposite angles.

Question 3. In figure, \( \angle PQR = 100^\circ \), where P, Q and R are points on a circle with centre O. Find \( \angle OPR \).
Answer: Take a point S in the major arc. Join PS and RS. Since PQRS is a cyclic quadrilateral, the sum of either pair of opposite angles is \( 180^\circ \).
\( \implies \angle PQR + \angle PSR = 180^\circ \)
\( \implies 100^\circ + \angle PSR = 180^\circ \)
\( \implies \angle PSR = 180^\circ - 100^\circ \)
\( \implies \angle PSR = 80^\circ \) .......(1)
Now, the angle formed by an arc at the centre is double the angle formed by it at any point on the remaining part of the circle.
\( \implies \angle POR = 2\angle PSR \)
\( \implies \angle POR = 2 \times 80^\circ \)
\( \implies \angle POR = 160^\circ \) .......(2)
In \( \triangle OPR \), \( OP = OR \) (radii of a circle).
Angles opposite to equal sides of a triangle are equal.
\( \implies \angle OPR = \angle ORP \) ............(3)
Also, the sum of all the angles of a triangle is \( 180^\circ \).
\( \implies \angle OPR + \angle ORP + \angle POR = 180^\circ \)
Using (2) and (3):
\( \implies \angle OPR + \angle OPR + 160^\circ = 180^\circ \)
\( \implies 2\angle OPR = 180^\circ - 160^\circ \)
\( \implies 2\angle OPR = 20^\circ \)
\( \implies \angle OPR = 10^\circ \)

In simple words: We first find the opposite angle in the cyclic quadrilateral (PQRS) to \( \angle PQR \). Then, we use the property that the central angle \( \angle POR \) is double the angle on the circumference \( \angle PSR \). Finally, because \( \triangle OPR \) is an isosceles triangle (two sides are radii), we can find \( \angle OPR \) using the sum of angles in a triangle.

Exam Tip: For problems involving cyclic quadrilaterals and central angles, remember two key theorems: opposite angles of a cyclic quadrilateral sum to \( 180^\circ \), and the angle at the center is double the angle at the circumference.

Question 4. In figure, \( \angle ABC = 69^\circ \), \( \angle ACB = 31^\circ \), find \( \angle BDC \).
Answer: In \( \triangle ABC \):
The sum of all the angles of a triangle is \( 180^\circ \).
\( \implies \angle BAC + \angle ABC + \angle ACB = 180^\circ \)
\( \implies \angle BAC + 69^\circ + 31^\circ = 180^\circ \)
\( \implies \angle BAC + 100^\circ = 180^\circ \)
\( \implies \angle BAC = 180^\circ - 100^\circ \)
\( \implies \angle BAC = 80^\circ \) .......(1)
Now, angles in the same segment of a circle are equal.
\( \implies \angle BDC = \angle BAC \)
\( \implies \angle BDC = 80^\circ \) [Using (1)]

In simple words: First, we use the fact that the angles inside any triangle add up to \( 180^\circ \) to find \( \angle BAC \). Then, we use a circle property that angles made by the same arc in the same segment of a circle are equal. This means \( \angle BDC \) will be the same as \( \angle BAC \).

Exam Tip: Always calculate the unknown angle in the triangle first. Then, apply the "angles in the same segment" theorem, which is a common trick in circle geometry problems.

Question 5. In figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that \( \angle BEC = 130^\circ \) and \( \angle ECD = 20^\circ \). Find \( \angle BAC \).
Answer: In \( \triangle CDE \):
The exterior angle property of a triangle states that the exterior angle is equal to the sum of the two opposite interior angles.
\( \implies \angle BEC = \angle CDE + \angle DCE \)
\( \implies 130^\circ = \angle CDE + 20^\circ \)
\( \implies \angle CDE = 130^\circ - 20^\circ \)
\( \implies \angle CDE = 110^\circ \) .......(2)
Now, angles in the same segment of a circle are equal.
\( \implies \angle BAC = \angle CDE \)
\( \implies \angle BAC = 110^\circ \)

In simple words: First, use the property that an exterior angle of a triangle equals the sum of its two opposite interior angles to find \( \angle CDE \). Then, use the circle theorem that angles made by the same arc in the same segment of a circle are equal. This means \( \angle BAC \) will be the same as \( \angle CDE \).

Exam Tip: When dealing with intersecting chords or diagonals inside a circle, look for exterior angles of triangles formed by the intersections. Then, connect those to angles in the same segment.

Question 6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If \( \angle DBC = 70^\circ \), \( \angle BAC \) is \( 30^\circ \), find \( \angle BCD \). Further, If AB = BC, find \( \angle ECD \).
Answer: Angles in the same segment of a circle are equal.
\( \implies \angle CDB = \angle BAC \)
\( \implies \angle CDB = 30^\circ \) .......(1)
Given \( \angle DBC = 70^\circ \) .......(2)
In \( \triangle BCD \):
The sum of all the angles of a triangle is \( 180^\circ \).
\( \implies \angle BCD + \angle DBC + \angle CDB = 180^\circ \)
\( \implies \angle BCD + 70^\circ + 30^\circ = 180^\circ \) [Using (1) and (2)]
\( \implies \angle BCD + 100^\circ = 180^\circ \)
\( \implies \angle BCD = 180^\circ - 100^\circ \)
\( \implies \angle BCD = 80^\circ \) .......(3)
In \( \triangle ABC \), \( AB = BC \).
Angles opposite to equal sides of a triangle are equal.
\( \implies \angle BCA = \angle BAC \)
\( \implies \angle BCA = 30^\circ \) .......(4) (given)
Now, from (3): \( \angle BCD = 80^\circ \)
We also know that \( \angle BCD = \angle BCA + \angle ECD \).
\( \implies 80^\circ = 30^\circ + \angle ECD \)
\( \implies \angle ECD = 80^\circ - 30^\circ \)
\( \implies \angle ECD = 50^\circ \)

In simple words: First, we use the property that angles made by the same arc are equal to find \( \angle CDB \). Then, using the sum of angles in \( \triangle BCD \), we calculate \( \angle BCD \). If two sides of \( \triangle ABC \) are equal, then the angles opposite those sides are also equal. Finally, we break down \( \angle BCD \) into two parts to find \( \angle ECD \).

Exam Tip: Break down the problem into smaller parts: first find \( \angle BCD \) using triangle properties and angles in the same segment, then use the isosceles triangle property to find the last part, \( \angle ECD \).

Question 7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Answer: In \( \triangle OAB \) and \( \triangle OCD \):
\( OA = OC \) (Radii of a circle)
\( OB = OD \) (Radii of a circle)
\( \angle AOB = \angle COD \) (Vertically opposite angles)
Thus, \( \triangle OAB \cong \triangle OCD \) (SAS rule)
\( \implies AB = CD \) .......(1) (Corresponding Parts of Congruent Triangles are equal)
Similarly, we can show that \( AD = CB \) ........(2)
Since the opposite sides of quadrilateral ABCD are equal, it is a parallelogram.
Now, diagonal BD is also a diameter.
The angle in a semicircle is a right angle.
\( \implies \angle BAD = \frac {1}{2}\angle BOD \)
\( \implies \angle BAD = \frac {1}{2} \times 180^\circ \)
\( \implies \angle BAD = 90^\circ \)
Therefore, ABCD is a rectangle (A parallelogram with an angle of \( 90^\circ \) is a rectangle).

In simple words: First, we prove that the quadrilateral has opposite sides equal by using congruent triangles formed at the center. This shows it's a parallelogram. Then, since the diagonals are diameters, any angle formed in a semicircle is a right angle. This means one angle of the parallelogram is \( 90^\circ \), which makes it a rectangle.

Exam Tip: To prove a figure is a rectangle, first prove it's a parallelogram, then show that one of its angles is \( 90^\circ \). The "angle in a semicircle" theorem is key for the \( 90^\circ \) part.

Question 8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic. Prove that an isosceles trapezium is cyclic.
Answer: Given: ABCD is a trapezium whose two non-parallel sides AD and BC are equal. To Prove: Trapezium ABCD is cyclic.
Proof: Draw BE parallel to AD from B to meet CD at E. (Construction)
Since AB is parallel to DE (given) and AD is parallel to BE (by construction), quadrilateral ABED is a parallelogram.
\( \implies \angle BAD = \angle BED \) .......(1) (Opposite angles of a parallelogram)
And \( AD = BE \) (Opposite sides of a parallelogram)
But it is given that \( AD = BC \) .........(3)
From (2) and (3), \( BE = BC \).
In \( \triangle BEC \), since \( BE = BC \), it is an isosceles triangle.
Angles opposite to equal sides are equal.
\( \implies \angle BEC = \angle BCE \) .........(4)
\( \angle BEC + \angle BED = 180^\circ \) (Linear Pair Axiom)
\( \implies \angle BCE + \angle BAD = 180^\circ \) [From (4) and (1)]
Therefore, Trapezium ABCD is cyclic. (If a pair of opposite angles of a quadrilateral sums to \( 180^\circ \), then the quadrilateral is cyclic.)

In simple words: To prove an isosceles trapezium is cyclic, we first draw a line parallel to one of its non-parallel sides, forming a parallelogram. This helps us show that some angles are equal. Then, using properties of isosceles triangles and linear pairs, we prove that a pair of opposite angles in the trapezium add up to \( 180^\circ \), which is the condition for a quadrilateral to be cyclic.

Exam Tip: For proofs involving trapeziums, often construction of a parallel line to form a parallelogram or a triangle is useful. Remember that a cyclic quadrilateral has opposite angles supplementary.

Question 9. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see figure). Prove that \( \angle ACP = \angle QCD \).
Answer: Angles in the same segment of a circle are equal.
\( \implies \angle ACP = \angle ABP \) ......(1)
And \( \angle QCD = \angle QBD \) ............(2)
Vertically opposite angles are equal.
\( \implies \angle ABP = \angle QBD \)
From (1) and (2), and the fact that \( \angle ABP = \angle QBD \), we can conclude:
\( \implies \angle ACP = \angle QCD \)

In simple words: This proof uses the theorem that angles formed by the same arc in the same segment of a circle are equal. We apply this to both circles to relate \( \angle ACP \) to \( \angle ABP \) and \( \angle QCD \) to \( \angle QBD \). Then, because \( \angle ABP \) and \( \angle QBD \) are vertically opposite angles and therefore equal, we can conclude that \( \angle ACP \) is equal to \( \angle QCD \).

Exam Tip: For problems involving intersecting circles and lines, always identify angles in the same segment and vertically opposite angles. These properties are key to connecting angles across different parts of the figure.

Question 10. If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lies on the third side.
Answer: Given: Circles are described with sides AB and AC of a triangle ABC as diameters. They intersect at a point D.
To Prove: D lies on the third side BC of \( \triangle ABC \).
Construction: Join AD.
Proof: For the circle described on AB as diameter, point D lies on the circumference.
The angle in a semicircle is a right angle.
\( \implies \angle ADB = 90^\circ \) .......(1)
Similarly, for the circle described on AC as diameter, point D lies on the circumference.
\( \implies \angle ADC = 90^\circ \) ...........(2)
Now, adding (1) and (2) we get:
\( \angle ADB + \angle ADC = 90^\circ + 90^\circ \)
\( \implies \angle ADB + \angle ADC = 180^\circ \)
Since the sum of these two adjacent angles is \( 180^\circ \), points B, D, C are collinear.
Therefore, D lies on BC.

In simple words: When two circles are drawn using two sides of a triangle as diameters, they will meet at a point on the third side. This is because the angle in a semicircle is always a right angle \( (90^\circ) \). If we connect the intersection point to the vertex A, both angles \( \angle ADB \) and \( \angle ADC \) are \( 90^\circ \). Since they add up to \( 180^\circ \), points B, D, and C must lie on a straight line.

Exam Tip: This theorem relies entirely on the property that an angle subtended by a diameter at any point on the circumference is \( 90^\circ \). Clearly state this property and show how the two right angles add up to \( 180^\circ \) to prove collinearity.

Question 11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that \( \angle CAD = \angle CBD \).
Answer: Given: ABC and ADC are two right triangles with common hypotenuse AC.
Since \( \triangle ABC \) is a right triangle with hypotenuse AC, it means \( \angle ABC = 90^\circ \). All vertices A, B, C lie on a circle where AC is the diameter. Similarly, since \( \triangle ADC \) is a right triangle with hypotenuse AC, it means \( \angle ADC = 90^\circ \). All vertices A, D, C also lie on a circle where AC is the diameter. Therefore, points A, B, C, and D are concyclic (lie on the same circle) with AC as its diameter.
Now, in both cases (whether B and D are on the same side or opposite sides of AC), DC is a chord of this circle.
Angles in the same segment of a circle are equal.
\( \implies \angle CAD = \angle CBD \)

In simple words: Since both triangles ABC and ADC are right-angled with the same hypotenuse AC, all four points A, B, C, and D lie on a single circle where AC is the diameter. Because these points are concyclic, angles \( \angle CAD \) and \( \angle CBD \) are angles in the same segment (formed by chord CD), making them equal.

Exam Tip: A key property to remember is that if two right-angled triangles share a common hypotenuse, all four vertices are concyclic, and the common hypotenuse becomes the diameter of the circumcircle. This allows you to apply theorems related to angles in the same segment.

Question 12. Prove that a cyclic parallelogram is a rectangle.
Answer: Given: ABCD is a cyclic parallelogram.
To prove: ABCD is a rectangle.
Proof: Since ABCD is a cyclic quadrilateral, its opposite angles are supplementary.
\( \implies \angle A + \angle C = 180^\circ \) .......(1)
Since ABCD is a parallelogram, its opposite angles are equal.
\( \implies \angle A = \angle C \) .........(2)
From (1) and (2):
\( \angle A + \angle A = 180^\circ \)
\( \implies 2\angle A = 180^\circ \)
\( \implies \angle A = 90^\circ \)
A parallelogram with one angle equal to \( 90^\circ \) is a rectangle.
Therefore, parallelogram ABCD is a rectangle.

In simple words: A cyclic parallelogram must be a rectangle. This is because in a cyclic quadrilateral, opposite angles add up to \( 180^\circ \). In a parallelogram, opposite angles are also equal. Combining these two facts, we find that each angle must be \( 90^\circ \). A parallelogram with a right angle is, by definition, a rectangle.

Exam Tip: This is a standard proof. Remember the two key properties: opposite angles of a cyclic quadrilateral are supplementary, and opposite angles of a parallelogram are equal. Combining these will directly lead to the conclusion that each angle is \( 90^\circ \).