Maharashtra Board Class 10 Maths Chapter 3 Circle Set 3.4 Solutions

Question 1. In the adjoining figure, in a circle with centre O, length of chord AB is equal to the radius of the circle. Find measure of each of the following.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसका केंद्र O है, और एक जीवा AB है जिसकी लम्बाई वृत्त की त्रिज्या के बराबर है। वृत्त पर बिंदु C स्थित है। इस चित्र में, वृत्त केंद्र O, जीवा AB, और वृत्त पर स्थित बिंदु C को दर्शाया गया है। जीवा AB की लंबाई त्रिज्या के बराबर होने के कारण, त्रिभुज OAB एक समबाहु त्रिभुज है।
Answer: (i) seg OA = seg OB = radius...... (i) [Radii of the same circle] seg AB = radius...... (ii) [Given] Therefore, seg OA = seg OB = seg AB [From (i) and (ii)] Therefore, \(\triangle OAB\) is an equilateral triangle. Therefore, \(m\angle AOB = 60^\circ\) [Angle of an equilateral triangle] (ii) \(m \angle ACB = \frac{1}{2} m \angle AOB\) [Measure of an angle subtended by an arc at a point on the circle is half of the measure of the angle subtended by the arc at the centre] \( = \frac{1}{2} \times 60^\circ\) Therefore, \(m \angle ACB = 30^\circ\) (iii) \(m(arc AB) = m \angle AOB\) [Definition of measure of minor arc] Therefore, \(m(arc AB) = 60^\circ\) (iv) \(m(arc ACB) + m(arc AB) = 360^\circ\) [Measure of a circle is \(360^\circ\)] Therefore, \(m(arc ACB) = 360^\circ - m(arc AB)\) \( = 360^\circ - 60^\circ\) Therefore, \(m(arc ACB) = 300^\circ\) In simple words: Given that chord AB equals the radius, triangle AOB is equilateral, making angle AOB \(60^\circ\). Angle ACB is half of AOB, so it's \(30^\circ\). Arc AB is equal to angle AOB, which is \(60^\circ\). Finally, arc ACB is the remaining part of the circle, \(360^\circ - 60^\circ = 300^\circ\).

🎯 Exam Tip: Remember the relationship between central angles, inscribed angles, and their intercepted arcs, especially in equilateral triangles formed by radii and chords. Make sure to clearly state the reasons for each step.

Question 2. In the adjoining figure, JPQRS is cyclic, side PQ = side RQ, \(\angle PSR = 110^\circ\). Find
ℹ️ चित्र व्याख्या (Diagram Explanation): एक चक्रीय चतुर्भुज PQRS है जिसमें P, Q, R, S वृत्त पर स्थित बिंदु हैं। भुजा PQ, भुजा RQ के बराबर है और कोण PSR \(110^\circ\) दिया गया है। यह चित्र एक वृत्त के अंदर चक्रीय चतुर्भुज को दर्शाता है जहाँ PQ और RQ जीवाएँ हैं।
(i) measure of \(\angle PQR\)
(ii) m (arc PQR)
(iii) m (arc QR)
(iv) measure of \(\angle PRQ\)
Answer: (i) JPQRS is a cyclic quadrilateral. [Given] Therefore, \(\angle PSR + \angle PQR = 180^\circ\) [Opposite angles of a cyclic quadrilateral are supplementary] Therefore, \(110^\circ + \angle PQR = 180^\circ\) Therefore, \(\angle PQR = 180^\circ - 110^\circ\) Therefore, \(m \angle PQR = 70^\circ\) (ii) \(\angle PSR = \frac{1}{2} m (arc PQR)\) [Inscribed angle theorem] \(110^\circ = \frac{1}{2} m (arc PQR)\) Therefore, \(m(arc PQR) = 220^\circ\) (iii) In \(\triangle PQR\), side PQ = side RQ [Given] Therefore, \(\angle PRQ = \angle QPR\) [Isosceles triangle theorem] Let \(\angle PRQ = \angle QPR = x\) Now, \(\angle PQR + \angle QPR + \angle PRQ = 180^\circ\) [Sum of the measures of angles of a triangle is \(180^\circ\)] Therefore, \(70^\circ + x + x = 180^\circ\) Therefore, \(70^\circ + 2x = 180^\circ\) Therefore, \(2x = 180^\circ - 70^\circ\) Therefore, \(2x = 110^\circ\) \(x = \frac{110^\circ}{2} = 55^\circ\) Therefore, \(\angle PRQ = \angle QPR = 55^\circ\)..... (i) But, \(\angle QPR = \frac{1}{2} m(arc QR)\) [Inscribed angle theorem] Therefore, \(55^\circ = \frac{1}{2} m(arc QR)\) Therefore, \(m(arc QR) = 110^\circ\) (iv) \(\angle PRQ = \angle QPR = 55^\circ\) [From (i)] Therefore, \(m \angle PRQ = 55^\circ\) In simple words: For the cyclic quadrilateral PQRS, opposite angles sum to \(180^\circ\), so \(\angle PQR = 180^\circ - 110^\circ = 70^\circ\). Arc PQR is twice the inscribed angle \(\angle PSR\), so \(220^\circ\). Since PQ=RQ, \(\triangle PQR\) is isosceles, making \(\angle PRQ = \angle QPR\). Using the sum of angles in a triangle, these angles are \(55^\circ\). Arc QR is twice \(\angle QPR\), which is \(110^\circ\). Finally, \(\angle PRQ\) is \(55^\circ\).

🎯 Exam Tip: When dealing with cyclic quadrilaterals, remember properties like opposite angles being supplementary. For inscribed angles and arcs, the relationship \(Angle = \frac{1}{2} Arc\) is crucial. For isosceles triangles, remember base angles are equal.

Question 3. MRPN is cyclic, \(\angle R = (5x -13)^\circ\), \(\angle N = (4x + 4)^\circ\). Find measures of \(\angle R\) and \(\angle N\).
Answer: MRPN is a cyclic quadrilateral. [Given] Therefore, \(\angle R + \angle N = 180^\circ\) [Opposite angles of a cyclic quadrilateral are supplementary]
ℹ️ चित्र व्याख्या (Diagram Explanation): एक चक्रीय चतुर्भुज MRPN को दर्शाता है, जिसके शीर्ष M, R, P, N एक वृत्त पर स्थित हैं। यह चित्र कोणों R और N के बीच के संबंध को समझने में मदद करता है। Therefore, \(5x - 13 + 4x + 4 = 180\) Therefore, \(9x - 9 = 180\) Therefore, \(9x = 189\) Therefore, \(x = \frac{189}{9}\) Therefore, \(x = 21\) Therefore, \(\angle R = 5x - 13\) \( = 5 \times 21 - 13\) \( = 105 - 13\) \( = 92^\circ\) \(\angle N = 4x + 4\) \( = 4 \times 21 + 4\) \( = 84 + 4\) \( = 88^\circ\) Therefore, \(m\angle R = 92^\circ\) and \(m \angle N = 88^\circ\) In simple words: Since MRPN is a cyclic quadrilateral, its opposite angles \(\angle R\) and \(\angle N\) are supplementary. By setting their sum to \(180^\circ\) and solving for \(x\), we find \(x=21\). Substituting \(x\) back into the expressions for \(\angle R\) and \(\angle N\) gives their measures as \(92^\circ\) and \(88^\circ\) respectively.

🎯 Exam Tip: The key property of cyclic quadrilaterals is that opposite angles are supplementary (sum to \(180^\circ\)). Use this to form an equation and solve for unknown variables, then substitute back to find the angle measures.

Question 4. In the adjoining figure, seg RS is a diameter of the circle with centre O. Point T lies in the exterior of the circle. Prove that \(\angle RTS\) is an acute angle.
Answer: Given: O is the centre of the circle, seg RS is the diameter of the circle. To prove: \(\angle RTS\) is an acute angle. Construction: Let seg RT intersect the circle at point P. Join PS and PT.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त को दिखाया गया है जिसका केंद्र O है और RS उसका व्यास है। बिंदु P व्यास RS पर स्थित है। बिंदु T वृत्त के बाहर स्थित है, और RT वृत्त को P पर प्रतिच्छेद करती है। यह आकृति कोण RTS को एक न्यून कोण सिद्ध करने के लिए एक ज्यामितीय निर्माण को दर्शाती है। Proof: seg RS is the diameter. [Given] Therefore, \(\angle RPS = 90^\circ\) [Angle inscribed in a semicircle] Now, \(\angle RPS\) is the exterior angle of \(\triangle PTS\). Therefore, \(\angle RPS > \angle PTS\) [Exterior angle is greater than the remote interior angles] Therefore, \(90^\circ > \angle PTS\) i.e. \(\angle PTS < 90^\circ\) i.e, \(\angle RTS < 90^\circ\) [R - P -T] \(\angle RTS\) is an acute angle. In simple words: Since RS is a diameter, the angle \(\angle RPS\) inscribed in the semicircle is \(90^\circ\). For triangle PTS, \(\angle RPS\) is an exterior angle, which means it must be greater than the interior remote angle \(\angle PTS\). Therefore, \(\angle PTS\) (which is the same as \(\angle RTS\)) must be less than \(90^\circ\), proving it is an acute angle.

🎯 Exam Tip: Key concepts here are the angle inscribed in a semicircle (\(90^\circ\)) and the exterior angle property of a triangle (exterior angle > remote interior angle). Combining these two is essential to prove the angle is acute.

Question 5. Prove that, any rectangle is a cyclic quadrilateral.
Answer: Given: \(\square ABCD\) is a rectangle. To prove: \(\square ABCD\) is a cyclic quadrilateral.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक आयत ABCD को दर्शाया गया है जिसके शीर्ष A, B, C, D हैं। यह चित्र यह सिद्ध करने के लिए है कि एक आयत एक चक्रीय चतुर्भुज होता है, जहाँ आयत के सभी कोण \(90^\circ\) होते हैं। Proof: \(\square ABCD\) is a rectangle. [Given] Therefore, \(\angle A = \angle B = \angle C = \angle D = 90^\circ\) [Angles of a rectangle] Now, \(\angle A + \angle C = 90^\circ + 90^\circ\) Therefore, \(\angle A + \angle C = 180^\circ\) Therefore, \(\square ABCD\) is a cyclic quadrilateral. [Converse of cyclic quadrilateral theorem] In simple words: All angles in a rectangle are \(90^\circ\). If we take any pair of opposite angles, such as \(\angle A\) and \(\angle C\), their sum is \(90^\circ + 90^\circ = 180^\circ\). Since a pair of opposite angles is supplementary, by the converse of the cyclic quadrilateral theorem, the rectangle must be a cyclic quadrilateral.

🎯 Exam Tip: To prove a quadrilateral is cyclic, you need to show that a pair of its opposite angles are supplementary. For a rectangle, all angles are \(90^\circ\), making this proof straightforward.

Question 6. In the adjoining figure, altitudes YZ and XT of \(\triangle WXY\) intersect at P. Prove that,
Answer:
(i) \(\square WZPT\) is cyclic.
(ii) Points X, Z, T, Y are concyclic. Given: seg YZ \(\perp\) side XW seg XT \(\perp\) side WY To prove: i. \(\square WZPT\) is cyclic. ii. Points X, Z, T, Y are concyclic.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक त्रिभुज WXY को दर्शाया गया है जहाँ शीर्ष W से XT और Y से YZ ऊंचाईयां हैं, जो बिंदु P पर प्रतिच्छेद करती हैं। यह चित्र त्रिभुज में ऊँचाईयों के प्रतिच्छेदन बिंदु से बने चक्रीय चतुर्भुज और समचक्रीय बिंदुओं को सिद्ध करने में मदद करता है। Proof: i. seg YZ \(\perp\) side XW [Given] Therefore, \(\angle PZW = 90^\circ\) ...... (i) seg XT \(\perp\) side WY [Given] Therefore, \(\angle PTW = 90^\circ\) ......(ii) \(\angle PZW + \angle PTW = 90^\circ + 90^\circ\) [Adding (i) and (ii)] Therefore, \(\angle PZW + \angle PTW = 180^\circ\) \(\square WZPT\) is a cyclic quadrilateral. [Converse of cyclic quadrilateral theorem] ii. \(\angle XZY = \angle YTX = 90^\circ\) [Given] Therefore, Points X and Y on line XY subtend equal angles on the same side of line XY. Therefore, Points X, Z, T and Y are concyclic. [If two points on a given line subtend equal angles at two distinct points which lie on the same side of the line, then the four points are concyclic] In simple words: For part (i), since YZ and XT are altitudes, angles \(\angle PZW\) and \(\angle PTW\) are both \(90^\circ\). Their sum is \(180^\circ\), making \(\square WZPT\) a cyclic quadrilateral by the converse theorem. For part (ii), because \(\angle XZY\) and \(\angle YTX\) are both \(90^\circ\) and subtend the same segment XY, points X, Z, T, Y are concyclic.

🎯 Exam Tip: For proving a quadrilateral is cyclic, look for opposite angles summing to \(180^\circ\) or two points subtending equal angles on the same side of a line. Altitudes always create \(90^\circ\) angles, which are often key for these proofs.

Question 7. In the adjoining figure, \(m (arc NS) = 125^\circ\), \(m(arc EF) = 37^\circ\), find the measure of \(\angle NMS\).
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त को दर्शाया गया है जिसमें दो जीवाएँ EN और FS वृत्त के बाहर बिंदु M पर प्रतिच्छेद करती हैं। चाप NS और चाप EF के माप दिए गए हैं। यह चित्र वृत्त के बाहर प्रतिच्छेद करने वाली जीवाओं द्वारा बनने वाले कोण को दर्शाने के लिए है। Chords EN and FS intersect externally at point M. \(m\angle NMS = \frac{1}{2} [m (arc NS) - m(arc EF)]\) \( = \frac{1}{2} (125^\circ - 37^\circ)\) \( = \frac{1}{2} \times 88^\circ\) Therefore, \(m\angle NMS = 44^\circ\) In simple words: When two chords intersect outside a circle, the measure of the angle formed is half the difference of the measures of the intercepted arcs. Given arc NS is \(125^\circ\) and arc EF is \(37^\circ\), \(\angle NMS\) is calculated as half of \((125^\circ - 37^\circ)\), which simplifies to \(44^\circ\).

🎯 Exam Tip: Remember the formula for the angle formed by two chords intersecting *outside* a circle: \(Angle = \frac{1}{2} (Larger Arc - Smaller Arc)\). Distinguish this from the formula for chords intersecting *inside* the circle.

Question 8. In the adjoining figure, chords AC and DE intersect at B. If \(\angle ABE = 108^\circ\), \(m(arc AE) = 95^\circ\), find \(m (arc DC)\).
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त को दर्शाया गया है जिसमें दो जीवाएँ AC और DE वृत्त के भीतर बिंदु B पर प्रतिच्छेद करती हैं। कोण ABE और चाप AE के माप दिए गए हैं। यह चित्र वृत्त के अंदर प्रतिच्छेद करने वाली जीवाओं द्वारा बनने वाले कोण और संबंधित चापों को दर्शाता है। Solution: Chords AC and DE intersect internally at point B. Therefore, \(\angle ABE = \frac{1}{2} [m(arc AE) + m(arc DC)]\) Therefore, \(108^\circ = \frac{1}{2} [95^\circ + m(arc DC)]\) Therefore, \(108^\circ \times 2 = 95^\circ + m(arc DC)\) Therefore, \(95^\circ + m(arc DC) = 216^\circ\) Therefore, \(m(arc DC) = 216^\circ - 95^\circ\) Therefore, \(m(arc DC) = 121^\circ\) In simple words: When two chords intersect inside a circle, the angle formed is half the sum of the measures of the intercepted arcs. Using the given \(\angle ABE = 108^\circ\) and \(m(arc AE) = 95^\circ\), we can set up an equation to solve for \(m(arc DC)\), which comes out to be \(121^\circ\).

🎯 Exam Tip: For chords intersecting *inside* a circle, the angle is half the *sum* of the intercepted arcs. Be careful not to confuse this with the formula for chords intersecting *outside* the circle.

Question 1. Draw a sufficiently large circle of any radius as shown in the figure below. Draw a chord AB and central \(\angle ACB\). Take any point D on the major arc and point E on the minor arc.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसमें जीवा AB है और केंद्र C से \(\angle ACB\) बना है। बिंदु D दीर्घ चाप पर और बिंदु E लघु चाप पर स्थित है। वृत्त पर बिंदु F, G, H भी दीर्घ चाप ADB पर और बिंदु I लघु चाप AEB पर हैं। यह चित्र वृत्त के विभिन्न चापों और उन पर बनने वाले कोणों के संबंधों को दर्शाता है।
(i) Measure \(\angle ADB\) and \(\angle ACB\) and compare the measures.
(ii) Measure \(\angle ADB\) and \(\angle AEB\). Add the measures.
(iii) Take points F, G, H on the arc ADB. Measure \(\angle AFB\), \(\angle AGB\), \(\angle AHB\). Compare these measures with each other as well as with measure of \(\angle ADB\).
(iv) Take any point I on the arc AEB. Measure \(\angle AIB\) and compare it with \(\angle AEB\). (Textbook pg, no. 64)
Answer: i. \(\angle ACB = 2 \angle ADB\). ii. \(\angle ADB + \angle AEB = 180^\circ\). iii. \(\angle AHB = \angle ADB = \angle AFB = \angle AGB\) iv. \(\angle AEB = \angle AIB\) In simple words: This activity demonstrates key circle theorems: the central angle is double the inscribed angle subtending the same arc; opposite angles of a cyclic quadrilateral (formed by A,D,B,E) are supplementary; angles subtended by the same arc on the circle are equal.

🎯 Exam Tip: Understanding these fundamental circle theorems (central angle, inscribed angle, angles in the same segment, and properties of cyclic quadrilaterals) is crucial for solving geometry problems. Practical drawing helps visualize these relationships.

Question 2. Draw a sufficiently large circle with centre C as shown in the figure. Draw any diameter PQ. Now take points R, S, T on both the semicircles. Measure \(\angle PRQ\), \(\angle PSQ\), \(\angle PTQ\). What do you observe? (Textbook pg. no.65)
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसका केंद्र C है और PQ उसका व्यास है। बिंदु R, S, T वृत्त पर स्थित हैं, जिनमें से R और S एक अर्धवृत्त पर और T दूसरे अर्धवृत्त पर है। यह आकृति अर्धवृत्त में बने कोणों के माप को समझने के लिए है। \(\angle PRQ = \angle PSQ = \angle PTQ = 90^\circ\) [Student should draw and verily the above answers.] In simple words: When points R, S, and T are taken on a semicircle with diameter PQ, the angles \(\angle PRQ\), \(\angle PSQ\), and \(\angle PTQ\) will all measure \(90^\circ\). This illustrates the theorem that an angle inscribed in a semicircle is always a right angle.

🎯 Exam Tip: The theorem "Angle inscribed in a semicircle is a right angle (\(90^\circ\))" is very important. Recognize when a chord is a diameter, as any angle subtended by it on the circumference will be \(90^\circ\).

Question 3. Prove that, if two lines containing chords of a circle intersect each other outside the circle, then the measure of angle between them is half the difference in measures of the arcs intercepted by the angle. (Textbook pg. no. 72)
Answer: Given: Chord AB and chord CD intersect at E in the exterior of the circle. To prove: \(\angle AEC = \frac{1}{2} [m(arc AC) - m(arc BD)]\) Construction: Draw seg AD.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसमें दो जीवाएँ AB और CD वृत्त के बाहर बिंदु E पर प्रतिच्छेद करती हैं। यह चित्र वृत्त के बाहर प्रतिच्छेद करने वाली जीवाओं द्वारा बनने वाले कोण और उनके द्वारा अंतरित चापों के माप के बीच संबंध को सिद्ध करने के लिए एक ज्यामितीय निर्माण को दर्शाता है। Proof: \(\angle ADC\) is the exterior angle of \(\triangle ADE\). Therefore, \(\angle ADC = \angle DAE + \angle AED\) [Remote interior angle theorem] Therefore, \(\angle ADC = \angle DAE + \angle AEC\) [C - D - E] Therefore, \(\angle AEC = \angle ADC - \angle DAE\) ......(i) \(\angle ADC = \frac{1}{2} m(arc AC)\) (ii) [Inscribed angle theorem] \(\angle DAE = \frac{1}{2} m(arc BD)\) (iii) [A - B - E, Inscribed angle theorem] Therefore, \(\angle AEC = \frac{1}{2} m(arc AC) - \frac{1}{2} m (arc BD)\) [From (i), (ii) and (iii)] Therefore, \(\angle AEC = \frac{1}{2} [m(arc AC) - m (arc BD)]\) In simple words: We want to prove that the angle formed by two chords intersecting outside a circle is half the difference of the intercepted arcs. By constructing AD and using the exterior angle theorem for triangle ADE, we show \(\angle AEC = \angle ADC - \angle DAE\). Since \(\angle ADC\) and \(\angle DAE\) are inscribed angles, they are half of their respective intercepted arcs (AC and BD). Substituting these values proves the theorem.

🎯 Exam Tip: When proving theorems involving angles formed by chords/secants intersecting outside a circle, construction of an auxiliary line (like AD here) is often necessary. The exterior angle theorem and inscribed angle theorem are crucial tools.

Question 4. Angles inscribed in the same arc are congruent. Write 'given' and 'to prove' with the help of the given figure. Think of the answers of the following questions and write the proof.
Answer:
(i) Which arc is intercepted by \(\angle PQR\)?
(ii) Which arc is intercepted by \(\angle PSR\)?
(iii) What is the relation between an inscribed angle and the arc intercepted by it? (Textbook: pg. no. 68)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसका केंद्र C है। वृत्त पर चार बिंदु P, Q, R, S हैं, और एक बिंदु T वृत्त पर है जो PR और QS के बीच में स्थित है। यह चित्र एक ही चाप PTR द्वारा अंतरित कोणों \(\angle PQR\) और \(\angle PSR\) के संबंध को दर्शाता है। Given: C is the centre of circle. \(\angle PQR\) and \(\angle PSR\) are inscribed in same arc PTR. To prove: \(\angle PQR = \angle PSR\) Proof: i. arc PTR is intercepted by \(\angle PQR\). ii. arc PTR is intercepted by \(\angle PSR\). iii. \(\angle PQR = \frac{1}{2} m(arc PTR)\), and (i) [inscribed angle theorem] \(\angle PSR = \frac{1}{2} m(arc PTR)\) (ii) [Inscribed angle theorem] Therefore, \(\angle PQR = \angle PSR\) [From (i) and (ii)] In simple words: The question asks to prove that angles inscribed in the same arc are congruent. Both \(\angle PQR\) and \(\angle PSR\) intercept the same arc PTR. According to the inscribed angle theorem, each angle is half the measure of its intercepted arc. Since both angles are half of the same arc, they must be equal.

🎯 Exam Tip: The theorem "Angles in the same segment of a circle are equal" is fundamental. Always identify the common intercepted arc when proving or applying this theorem.

Question 5. Angle inscribed in a semicircle is a right angle. With the help of given figure write 'given', 'to prove' and 'the proof. (Textbook pg. no. 68)
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त जिसका केंद्र M है और AXC उसका व्यास है। वृत्त पर बिंदु B स्थित है, जिससे \(\angle ABC\) बना है। यह चित्र यह सिद्ध करने के लिए है कि अर्धवृत्त में बना कोण एक समकोण होता है। Given: M is the centre of circle. \(\angle ABC\) is inscribed in arc ABC. Arcs ABC and AXC are semicircles. To prove: \(\angle ABC = 90^\circ\) Proof: \(\angle ABC = \frac{1}{2} m(arc AXC)\) (i) [Inscribed angle theorem] arc AXC is a semicircle. Therefore, \(m(arc AXC) = 180^\circ\) (ii) [Measure of semicircular arc is \(180^\circ\)] Therefore, \(\angle ABC = \frac{1}{2} \times 180^\circ\) Therefore, \(\angle ABC = 90^\circ\) [From (i) and (ii)] In simple words: Given that AXC is a semicircle, its measure is \(180^\circ\). The angle \(\angle ABC\) is an inscribed angle that intercepts arc AXC. By the inscribed angle theorem, the measure of \(\angle ABC\) is half the measure of arc AXC. Therefore, \(\angle ABC = \frac{1}{2} \times 180^\circ = 90^\circ\), proving it is a right angle.

🎯 Exam Tip: The definition of an inscribed angle and the measure of a semicircle (\(180^\circ\)) are the core components of this proof. Clearly state each step and its reason.

Question 6. Theorem: Opposite angles of a cyclic quadrilateral are supplementry. Fill in the blanks and complete the following proof. (Textbook pg. no. 68)
Answer: Given: \(\square ABCD\) is cyclic. To prove: \(\angle B + \angle D = 180^\circ\) \(\angle A + \angle C = 180^\circ\)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त के अंदर चक्रीय चतुर्भुज ABCD को दर्शाया गया है जिसके शीर्ष A, B, C, D वृत्त पर स्थित हैं। यह चित्र चक्रीय चतुर्भुज के सम्मुख कोणों के संपूरक होने के प्रमेय को सिद्ध करने के लिए है। Proof: arc ABC is intercepted by the inscribed angle \(\angle ADC\). Therefore, \(\angle ADC = \frac{1}{2} m(arc ABC)\) (i) [Inscribed angle theorem] Similarly, \(\angle ABC\) is an inscribed angle. It intercepts arc ADC. Therefore, \(\angle ABC = \frac{1}{2} m(arc ADC)\) (ii) [Inscribed angle theorem] Therefore, \(\angle ADC + \angle ABC = \frac{1}{2} m(arc ABC) + \frac{1}{2} m(arc ADC)\) [Adding (i) and (ii)] Therefore, \(\angle D + \angle B = \frac{1}{2} [m(arc ABC) + m(arc ADC)]\) Therefore, \(\angle B + \angle D = \frac{1}{2} \times 360^\circ\) [arc ABC and arc ADC constitute a complete circle] \( = 180^\circ\) Therefore, \(\angle B + \angle D = 180^\circ\) Similarly we can prove, \(\angle A + \angle C = 180^\circ\) In simple words: The proof shows that for a cyclic quadrilateral, the sum of opposite angles is \(180^\circ\). By applying the inscribed angle theorem, \(\angle ADC\) is half of arc ABC, and \(\angle ABC\) is half of arc ADC. Summing these two equations, we get that \(\angle ADC + \angle ABC\) is half the sum of arc ABC and arc ADC, which is half of the entire circle (\(360^\circ\)), resulting in \(180^\circ\).

🎯 Exam Tip: This is a key theorem for cyclic quadrilaterals. Understand that the sum of the two intercepted arcs by a pair of opposite angles forms the entire circle (\(360^\circ\)). The inscribed angle theorem then directly leads to their sum being \(180^\circ\).

Question 7. In the above theorem, after proving \(\angle B + \angle D = 180^\circ\), can you use another way to prove \(\angle A + \angle C = 180^\circ\)? (Textbook pg. no. 69)
Answer: Proof: Yes, we can prove \(\angle A + \angle C = 180^\circ\) by another way. \(\angle B + \angle D = 180^\circ\) In \(\square ABCD\), \(\angle A + \angle B + \angle C + \angle D = 360^\circ\) [Sum of the measures of all angles of a quadrilateral is \(360^\circ\).] Therefore, \(\angle A + \angle C + 180^\circ = 360^\circ\) Therefore, \(\angle A + \angle C = 360^\circ - 180^\circ\) Therefore, \(\angle A + \angle C = 180^\circ\) In simple words: Yes, if we already know that \(\angle B + \angle D = 180^\circ\) in a quadrilateral, and we also know that the sum of all interior angles in any quadrilateral is \(360^\circ\), then we can simply substitute the known sum of \(\angle B + \angle D\) into the total sum equation. This leaves us with \(\angle A + \angle C = 180^\circ\).

🎯 Exam Tip: This alternative proof highlights the usefulness of knowing the sum of interior angles of a quadrilateral (\(360^\circ\)). It's a faster way to deduce the supplementary nature of the second pair of opposite angles once the first pair is proven.

Question 8. An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle. (Textbook pg. no. 69)
Answer: Given: \(\square ABCD\) is a cyclic quadrilateral. \(\angle BCE\) is the exterior angle of \(\square ABCD\). To prove: \(\angle BCE = \angle BAD\)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक चक्रीय चतुर्भुज ABCD को दर्शाया गया है, जिसके शीर्ष A, B, C, D एक वृत्त पर स्थित हैं। भुजा DC को E तक बढ़ाया गया है, जिससे \(\angle BCE\) एक बाह्य कोण बनता है। यह चित्र चक्रीय चतुर्भुज के बाह्य कोण के गुण को दर्शाता है। Proof: \(\angle BCE + \angle BCD = 180^\circ\) ...... (i) [Angles in a linear pair] \(\square ABCD\) is a cyclic quadrilateral. [Given] \(\angle BAD + \angle BCD = 180^\circ\) .......... (ii) [Opposite angles of a cyclic quadrilateral are supplementary] Therefore, \(\angle BCE + \angle BCD = \angle BAD + \angle BCD\) [From (i) and (ii)] Therefore, \(\angle BCE = \angle BAD\) In simple words: An exterior angle (\(\angle BCE\)) and its adjacent interior angle (\(\angle BCD\)) form a linear pair, summing to \(180^\circ\). In a cyclic quadrilateral, opposite interior angles (\(\angle BAD\) and \(\angle BCD\)) also sum to \(180^\circ\). By equating these two sums, we can conclude that the exterior angle \(\angle BCE\) is equal to the interior opposite angle \(\angle BAD\).

🎯 Exam Tip: This theorem is a direct consequence of two key properties: angles in a linear pair sum to \(180^\circ\), and opposite angles of a cyclic quadrilateral sum to \(180^\circ\). Clearly stating these reasons is crucial.

Question 9. Theorem: If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic. (Textbook pg. no. 69)
Answer: Given: In \(\square ABCD\), \(\angle A + \angle C = 180^\circ\) To prove: \(\square ABCD\) is a cyclic quadrilateral. Proof: (Indirect method) Suppose \(\square ABCD\) is not a cyclic quadrilateral. We can still draw a circle passing through three non collinear points A, B, D.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त को दर्शाया गया है जो तीन असारेखीय बिंदुओं A, B, D से होकर गुजरता है। बिंदु C वृत्त के बाहर स्थित है, और एक काल्पनिक बिंदु E वृत्त पर स्थित है, जैसे कि D-E-C. यह चित्र एक अप्रत्यक्ष विधि से चक्रीय चतुर्भुज प्रमेय को सिद्ध करने के लिए है। Case I: Point C lies outside the circle. Then, take point E on the circle such that D - E - C. Therefore, \(\square ABED\) is a cyclic quadrilateral. \(\angle DAB + \angle DEB = 180^\circ\) (i) [Opposite angles of a cyclic quadrilateral are supplementary] \(\angle DAB + \angle DCB = 180^\circ\) (ii) [Given] Therefore, \(\angle DAB + \angle DEB = \angle DAB + \angle DCB\) [From (i) and (ii)] Therefore, \(\angle DEB = \angle DCB\)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त को दर्शाया गया है जो तीन बिंदुओं A, B, D से होकर गुजरता है। बिंदु C वृत्त के बाहर स्थित है, और एक बिंदु E वृत्त पर स्थित है, जैसे कि D-E-C. यह आकृति यह तर्क देने के लिए है कि \(\angle DEB\) और \(\angle DCB\) समान नहीं हो सकते, जिससे यह साबित होता है कि बिंदु C वृत्त के बाहर नहीं हो सकता। But, \(\angle DEB \neq \angle DCB\) as \(\angle DEB\) is the exterior angle of \(\triangle BEC\). Therefore, Our supposition is wrong. Therefore, \(\square ABCD\) is a cyclic quadrilateral. In simple words: This proof uses the indirect method. Assume \(\square ABCD\) is not cyclic. Then, a circle passes through A, B, D. If C is outside the circle, we construct a point E on the circle such that D-E-C. This makes \(\square ABED\) cyclic, so \(\angle DAB + \angle DEB = 180^\circ\). Given \(\angle DAB + \angle DCB = 180^\circ\), we deduce \(\angle DEB = \angle DCB\). However, \(\angle DEB\) is the exterior angle of \(\triangle BEC\) and should be greater than \(\angle DCB\), leading to a contradiction. Thus, C cannot be outside.

🎯 Exam Tip: Indirect proofs (proof by contradiction) are common in geometry. The strategy is to assume the opposite of what you want to prove and then derive a contradiction using known theorems. Be clear about the initial assumption and the contradiction reached.

Therefore, \(\square ABCD\) is a cyclic quadrilateral. Case II: Point C lies inside the circle. Then, take point E on the circle such that D-C-E Therefore, \(\square ABED\) is a cyclic quadrilateral. \(\angle DAB + \angle DEB = 180^\circ\) (iii) [Opposite angles of a cyclic quadrilateral are supplementary] \(\angle DAB + \angle DCB = 180^\circ\) (iv) [Given] Therefore, \(\angle DAB + \angle DEB = \angle DAB + \angle DCB\) [From (iii) and (iv)] Therefore, \(\angle DEB = \angle DCB\) But \(\angle DEB \neq \angle DCB\) as \(\angle DCB\) is the exterior angle of \(\triangle BCE\). Therefore, Our supposition is wrong. Therefore, \(\square ABCD\) is a cyclic quadrilateral. In simple words: In Case II, if C lies inside the circle, we extend DC to meet the circle at E, making D-C-E. \(\square ABED\) is cyclic, so \(\angle DAB + \angle DEB = 180^\circ\). With the given \(\angle DAB + \angle DCB = 180^\circ\), it implies \(\angle DEB = \angle DCB\). But \(\angle DCB\) is an exterior angle for \(\triangle BCE\), meaning it should be greater than \(\angle DEB\), which is a contradiction. Thus, C cannot be inside the circle either. Since C can neither be outside nor inside, it must lie on the circle, making \(\square ABCD\) cyclic.

🎯 Exam Tip: For the converse of the cyclic quadrilateral theorem, both cases (point outside and point inside) need to be disproven via contradiction. Make sure to clearly draw auxiliary lines and state the properties of exterior angles of a triangle.

Question 10. Theorem: If two points on a given line subtend equal angles at two distinct points which lie on the same side of the line, then the four points are concyclic. (Textbook pg. no. 70)
Answer: Given: Points B and C lie on the same side of the line AD. \(\angle ABD = \angle ACD\) To prove: Points A, B, C, D are concyclic. i.e., \(\square ABCD\) is a cyclic quadrilateral.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक रेखा AD है जिसके एक ही ओर दो बिंदु B और C हैं। \(\angle ABD\) और \(\angle ACD\) बराबर दिए गए हैं। यह चित्र यह सिद्ध करने के लिए है कि यदि दो बिंदु एक ही रेखाखंड पर समान कोण अंतरित करते हैं, तो वे चार बिंदु समचक्रीय होते हैं। Proof: Suppose points A, B, C, D are not concyclic points. We can still draw a circle passing through three non collinear points A, B, D. Case I: Point C lies outside the circle. Then, take point E on the circle such that A - E - C. \(\angle ABD = \angle AED\) (i) [Angles inscribed in the same arc] \(\angle ABD = \angle ACD\) (ii) [Given] Therefore, \(\angle AED = \angle ACD\) [From (i) and (ii)] In simple words: This theorem states that if two points (B and C) on the same side of a line segment (AD) subtend equal angles at the endpoints (A and D), then the four points (A, B, C, D) are concyclic. The proof uses contradiction: assuming C is outside the circle passing through A, B, D, we construct E on the circle. If \(\angle ABD = \angle ACD\) is given, and \(\angle ABD = \angle AED\) (angles in same segment), then \(\angle AED = \angle ACD\). This contradicts the property that an exterior angle of \(\triangle CDE\) (\(\angle AED\)) must be greater than its remote interior angle (\(\angle ACD\)).

🎯 Exam Tip: This is the converse of the "angles in the same segment" theorem. For indirect proofs, carefully set up the contradiction. The exterior angle property of a triangle is a frequent tool for creating these contradictions.

Therefore, \(\angle AED = \angle ECD\) [A - E - C]
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त को दर्शाया गया है जो तीन बिंदुओं A, B, D से होकर गुजरता है। बिंदु C वृत्त के बाहर स्थित है, और एक बिंदु E वृत्त पर स्थित है, जैसे कि A-E-C. यह आकृति यह तर्क देने के लिए है कि \(\angle AED\) और \(\angle ECD\) समान नहीं हो सकते, जिससे यह साबित होता है कि बिंदु C वृत्त के बाहर नहीं हो सकता। But, \(\angle AED \neq \angle ECD\) as \(\angle AED\) is the exterior angle of \(\triangle ECD\). Therefore, Our supposition is wrong. Therefore, Points A, B, C, D are concyclic points. Case II: Point C lies inside the circle. Then, take point E on the circle such that A - C - E. \(\angle ABD = \angle AED\) (iii) [Angles inscribed in the same arc] \(\angle ABD = \angle ACD\) (iv) [Given] Therefore, \(\angle AED = \angle ACD\) [From (iii) and (iv)] Therefore, \(\angle CED \simeq \angle ACD\) [A - C - E]
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त को दर्शाया गया है जो तीन बिंदुओं A, B, D से होकर गुजरता है। बिंदु C वृत्त के अंदर स्थित है, और एक बिंदु E वृत्त पर स्थित है, जैसे कि A-C-E. यह आकृति यह तर्क देने के लिए है कि \(\angle CED\) और \(\angle ACD\) समान नहीं हो सकते, जिससे यह साबित होता है कि बिंदु C वृत्त के अंदर नहीं हो सकता। But, \(\angle CED = \angle ACD\) as \(\angle ACD\) is the exterior angle of \(\triangle ECD\). Therefore, Our supposition is wrong. Therefore, Points A, B, C, D are concyclic points. In simple words: Case II examines if C lies inside the circle. By extending AC to meet the circle at E, we form \(\triangle CDE\). Given \(\angle ABD = \angle ACD\), and knowing \(\angle ABD = \angle AED\) (angles in the same segment), we get \(\angle AED = \angle ACD\). However, \(\angle ACD\) is an exterior angle to \(\triangle CDE\) and should be greater than \(\angle AED\), leading to a contradiction. Thus, C cannot be inside. Since C is neither outside nor inside, it must lie on the circle, making A, B, C, D concyclic.

🎯 Exam Tip: When using the indirect method, ensure that both possibilities (point outside and point inside) are rigorously disproven to fully establish the theorem. The geometric relationship between interior and exterior angles of a triangle is often the basis for the contradiction.

Question 11. The above theorem is converse of a certain theorem. State it. (Textbook pg. no. 70)
Answer: If four points are concyclic, then the line joining any two points subtend equal angles at the other two points which are on the same side of that line. In simple words: The stated theorem (Question 10) is the converse of "angles in the same segment are equal." The original theorem says if points are concyclic, angles subtended by the same segment are equal. The converse says if angles subtended by a segment are equal, then the points are concyclic.

🎯 Exam Tip: Understanding direct theorems and their converses is crucial. The converse essentially reverses the "if-then" statement. For this theorem, it's about establishing concyclicity from equal angles subtended by a common base.

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