**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.

**Solution:**

Given:- O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively.

Construction:- Join AC

Let us assumed that, ∠OAC = ∠OCA = x

∠AOC = 180˚ – 2x

∠BAC = 30˚+x

∠BCA = 40˚+x

In ΔABC

∠ABC = 180˚ – ∠BAC – ∠BCA

∠ABC = 180˚-(30˚+x) – (40˚ + x)

∠ABC = 180˚ – 30˚ – x – 40˚ – x

∠ABC = 110˚ – 2x

∠AOC = 2∠ABC (Angle at the centre is double the angle at the circumference)

180˚ – 2x = 2(110˚ – 2x)

2x = 40˚

x = 20˚

∠AOC = 180˚ – 2×40˚

∠AOC = 140˚

Hence, the ∠AOC is 140˚

**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.**

**Solution**

(i) Prove that AC is a diameter of the circle.

In ΔABD

∠DAB + ∠ABD + ∠ADB = 180˚

65˚ + 70˚ + ∠ADB = 180˚

∠ADB = 180˚-135˚

∠ADB = 45˚

∠ABC = ∠ADB + ∠BDC

∠ABC = 45˚ + 45˚

∠ABC = 90˚

Since, ∠ADC is the angle of semiarid,

Hence, AC is a diameter of the

(ii) We know that angles on the same segment of a circle are equal.

∠ACB = ∠ADB

∠ACB = 45˚

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

**Solution:**We have, ∠AOB = 2ACB (angle at the centre is double the angle at the circumference subtended by the same chord)

∠ACB = 70˚/2

∠ACB = 35˚

OC = OA (Radius of same circle)

∠OCA = ∠OAC = 35˚

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

(i)

(ii)

**Solution:**(i) We have b = 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 + 65˚ = 180˚

a = 180˚ – 65˚

a = 115˚

(ii) We have c = 1/2 × Reflex( 112˚) (angle at the centre is double the angle at the circumference subtended by the same chord)

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

**Solution:**

(i) ∠BAD = 90˚ (Angle in a semicircle)

∠BDA = 90˚ – 35˚

∠BDA = 55˚ (Angle subtended by the same chord on the circle are equal)

(ii) ∠DAC = ∠CBD = 25˚ (Angle subtended by the same chord on the circle are equal)

120˚ = b + 25˚

-b = 25˚-120˚

b = 95˚ (In a triangle measure of exterior angle is equal to the sum of pair of opposite interior angles)(iii) ∠AOB = 2∠AOB

∠AOB = 2×50˚

∠AOB = 100˚ (angle at the centre is double the angle at the circumference subtended by the same chord)

OA = OB

∠OBA = ∠OAB = c

c = (80˚-100˚/2

c = 80˚/2

c = 40˚

(iv) ∠AOB = 90˚ (angle in a semicircle)

∠BAP = 90˚-45˚

∠BAP = 45˚

d = ∠BCP = ∠BAP = 45˚ (Angle subtended by the same chord on the circle are equal).

**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. O_{1} and O_{2} are the centers of two circles.

**Solution:**∠DBA = 90˚ and ∠CBA = 90˚ (angle in a semicircle is a right angle)

∠DBC = 180˚

D, B and C form a straight line.

**Question **7. In the figure given below, find:

(i) ∠BCD

(ii) ∠ADC

(iii) ∠ABC

**Show steps of your working.****Solution:**

(i) ∠BCD + ∠BCD = 180˚ (Sum of opposite angle of a cyclic quadrilateral is 180˚)

∠BCD + 105˚ = 180˚

∠BCD = 180˚-105˚

∠BCD = 75˚

(ii) AB || CD

∠BAD + ∠ADC = 180˚ (Sum of opposite angle of a cyclic quadrilateral is 180˚)

105˚ + ∠ADC = 180˚

∠ADC = 180˚ – 105˚

∠ADC = 75˚

(iii) ∠ADC + ∠ABC = 180˚ (Sum of opposite angle of a cyclic quadrilateral is 180˚)

75˚ + ∠ABC = 180˚

∠ABC = 180˚ – 75˚

∠ADC = 105˚

**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**

**Solution:**

∠ACB = 1/2 Reflex (∠AOB) (Angle at the centre is double the angle at the circumference subtended by the same chord)

∠ACB = 1/2(360˚ – 140˚)

∠ACB = 1/2(220˚)

∠ACB = 110˚

OA = OB (Radius of same circle)

∠OBA = ∠OAB

= 180˚-140˚/2

= 40˚/2

= 20˚

∠CAB = 50˚-20˚

∠CAB = 30˚

In ΔCAB,

∠CBA = 180˚ – 110˚-30˚

∠CBA = 40˚

∠OBC = ∠CBA+∠OBA

∠OBC = 40˚+20˚

∠OBC = 60˚

**Question **9. Calculate:

(i) ∠CDB,

(ii) ∠ABC,

(ii) ∠ACB.

**Solution:**

It is given that,

∠CDB = ∠BAC = 49˚ (Angles subtended by the same chord on the circle are equal)

∠ABC = ∠ADC = 43˚ (Angles subtended by the same chord on the circle are equal)

By angle sum property we know that,

∠ACB = 180˚-49˚-43˚

∠ACB = 88˚

Hence, ∠ACB is 88˚.

**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

**Solution:**(i) By angle sum property of ΔABD,

∠ADB = 180˚ – 75˚ – 58˚

∠ADB = 47˚

∠BDC = ∠ADC – ∠ADB

∠BDC = 77˚-47˚

∠BDC = 30˚

(ii) ∠BAD + ∠BCD = 180˚ (Sum of opposite angles of a cyclic quadrilateral is 180˚)

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

**Question **11. In the following figure, O is centre of the circle and ∆ ABC is equilateral. Find:

(i) ∠ADB

(ii) ∠AEB

**Solution:**∠ACB and ∠ADB are in the same segment

So, ∠ADB = ∠ACB=60˚

Join OA and OB.

∠AOB = 2∠ACB

∠AOB = 2×60˚

∠AOB = 120˚

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

**Solution:**In ΔABC,

∠CBA = 50˚

∠CAB = 75˚

∠ACB = 180˚- (∠CBA + ∠CAB)

∠ACB = 180˚ – (∠CBA + ∠CAB)

∠ACB = 180˚ – (50˚ + 75˚)

∠ACB = 180˚ – 125˚

∠ACB = 55˚

∠ADB = ∠ACB = 55˚ (Angle subtended by the same chord on the circle are equal)

Again,

ΔABD,

∠DAB + ∠ABD + ∠ADB = 180˚

∠DAB + ∠ABD + 55˚= 180˚

∠DAB + ∠ABD = 180˚-55˚

∠DAB + ∠ABD = 125˚

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

**Solution:**We know that,

∠ACB = 90˚ (Angle in a semicircle is right angle)

∠ABC = 180˚ – ∠ADC

∠ABC = 180˚ – 130˚

∠ABC = 50˚ (Pair of opposite angles in a cyclic quadrilateral are supplementary)

∠BAC = 90˚ – ∠ABC

∠BAC = 90˚ – 50˚

∠BAC = 40˚ (by angle sum property of right triangle ACB)

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

**Solution:****Construction:- Join AD**

∠ADC = 1/2 ∠AOC

∠ADC = 1/2 × 110˚

∠ADC = 55˚(angle at the centre is double the angle at the circumference subtended by the same chord)

∠ADB = 90˚ (angle in a semicircle is a right angle)

∠BDC = 90˚ – ∠ADC

∠BDC = 90˚ – 55˚

∠BDC = 35˚

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

**Solution:**

Here,

∠ACB = 1/2 ∠AOB

∠ACB = 1/2 × 60˚

∠ACB = 30˚(angle at the centre is double the angle at the circumference subtended by the same chord)

ΔBDC, By angle sum property

∠DBC = 180˚-100˚-30˚

∠DBC = 50˚

Hence, the value of ∠OBC = 50˚

**Question **16. ABCD is a cyclic quadrilateral in which ∠ DAC = 27°; ∠ DBA = 50° and ∠ ADB = 33°.**Calculate:****(i) ∠DBC,****(ii) ∠DCB,****(iii) ∠CAB.**

**Solution:**(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)

∠ACB = ∠ADB = 50˚ (angles subtended by the same chord on the circle are equal)

∠DCB = ∠ACD + ∠ACB

∠DCB = 50˚ + 33˚

∠DCB = 83˚

(iii) ∠DAB + ∠DCB = 180˚ (pair of opposite angles in a cyclic quadrilateral are supplementary)

27˚ + ∠CAB + 83˚ = 180˚

∠CAB = 180˚ – 27˚ – 83˚

∠CAB = 180˚ – 110˚

∠CAB = 70˚

**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.

**Solution:**

(i) ∠CED = 90˚ (Angle in a semicircle is a right angle)

∠DCE = 90˚-∠CDE

∠DCE = 90˚-40˚

∠DCE = 50˚

∠DCE = ∠OCB = 50˚

(ii) In ΔBOC,

∠AOC = ∠OCB + ∠OBC (Exterior angle of a triangle is equal to the sum of pair of interior opposite angles)

80˚ = 50˚ + ∠OBC

∠OBC = 80-50˚

∠OBC = 30˚

**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.

**Solution:**

Construction:- Join OB.

OBA = 90˚ (Angle in a semicircle is a right angle)

OB ⏊ AE

We know, that perpendicular drawn from the centre to a chord bisects the chord.

AB = BE

Hence, AB=BE

**Question **19. In the following figure,

(i) if ∠BAD = 96°, find BCD and

(ii) Prove that AD is parallel to FE.

**Solution:**(i) ABCD is a cyclic quadrilateral

∠BAD + ∠BCD = 180˚(pair of opposite angles in a cyclic quadrilateral are supplementary)

∠BCD = 180˚-96˚

∠BCD = 84˚

∠BCE = 180˚-84˚

∠BCE = 96˚

Similarly,

BCEF is a cyclic quadrilateral

∠BCE + ∠BFE = 180˚ (pair of opposite angles in a cyclic quadrilateral are supplementary)

96˚ + ∠BFE = 180˚

∠BFE = 180˚ – 96˚

∠BFE = 84˚

(ii) ∠BAD + ∠BFE

= 96˚ + 84˚

= 180˚

These two are interior angles on the same side of a pair of lines AD and EF

AD‖FE┤

**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:

(i)

Let ABCD be a parallelogram inscribed in a circle.

∠BAD = ∠BCD (Opposite angle of a parallelogram are equal)

∠BAD + ∠BCD = 180˚ (Pair of opposite angle in a cyclic quadrilateral are supplementary)

∠BAD = ∠BCD = 180˚/2 = 90˚

Similarly, the other two angles are 90˚ and opposite pair of sides are equal.

Hence, ABCD is a rectangle.

(ii)

Let ABCD be a parallelogram inscribed in a circle.

∠BAD = ∠BCD (Opposite angle of a rhombus are equal)

∠BAD+∠BCD = 180˚ (Pair of opposite angle in a cyclic quadrilateral are supplementary)

∠BAD = ∠BCD = 180˚/2 = 90˚

Similarly, the other two angles are 90˚ and all the sides are equal.

Hence, ABCD is a square.

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

**Solution:**

AB = AC

∠B = ∠C

Hence, DECB is a cyclic quadrilateral

In a triangle angles opposite to equal sides are equal.

∠B + ∠DEC = 180˚_____ (i) (Pair of opposite angles in a cyclic quadrilateral are supplementary)

∠C + ∠DEC = 180˚ (this is the sum of interior angles on one side of a transversal)

DE || BC

∠ADE = ∠B (corresponding angles)

∠AED = ∠C (corresponding angles)

∠ADE = ∠AED

AD = AE

AB – AD = AC – AE (AB = AC)

BD = CE

We have DE || BC and BD = CE

Therefore, DECB is isosceles trapezium.

**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.

Solution:

Let O and O’ be the centers of two intersecting circles,

Where points of intersection are P and Q and PA and PB are their diameters respectively.

Construction:- Join PQ, AQ and QB.

∠AQP = 90˚

∠BQP = 90˚ (angle in a semicircle is a right angle adding both these angles)

∠AQP + ∠BQP = 180˚

∠AQB = 180˚

Therefore, the points A, Q and B are collinear.

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

**Solution:**

(i) ∠ABC = 1/2 Reflex ∠COA (Angle at the centre is double the angle at the circumference subtended by the same chord)

b = 1/2 (360˚ – a)

2b = 360˚ – a

a + 2b = 180˚

(ii) Since OABC is a parallelogram, so opposite angles are equal

a = b

Using relationship in (i),

3a = 180˚

a = 60˚

Also, OC||BA

∠COA + ∠OAB = 180˚

60˚ + ∠OAB = 180˚

∠OAB = 180˚- 60˚

∠OAB = 120˚

**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 as the centre O is equal to twice the angle APC.

Solution:

It is given that, two chords AB and CD intersect each other at P inside the circle. OA, OC and OD are joined.

Need to Prove: – ∠AOC + ∠BOD = 2∠APC

By Construction: – Join AD.

Proof:

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

∠AOC = 2∠ADC________(1)

Again,

∠BOD = 2∠BAD________(2)

By adding both the equations,

∠AOC + ∠BOD = 2∠ADC + 2∠BAD

∠AOC + ∠BOD = 2(∠ADC + ∠BAD)___________(3)

In ΔPAD,

∠APC = ∠PAD + ∠ADC

∠APC = ∠BAD + ∠ADC___________(4)

From equation (3) and (4)

∠AOC + ∠BOD = 2∠APC

Hence, the sum of the angles subtended by the arcs AC and BD as the centre O is equal to twice the angle APC.

**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.**

**Solution:**

(i) Join RN and MS

∠RMS = 90˚ (Angle in a semicircle is a right angle)

∠RMS = 90˚ – 29˚

∠RMS = 61˚

∠RNM = 180˚ – ∠RSM (by angle sum property of triangle RMS)

∠RNM = 180˚ – 61˚

∠RNM = 119˚ (Pair of opposite angles in a cyclic quadrilateral are supplementary).

(ii) Join RS||NM

∠NMR = ∠MRS = 29˚ (Alternate angles)

∠NMR = 90˚ + 29˚

∠NMR = 119˚

∠NRS + ∠NMS = 180˚ (Pair of opposite angles in cyclic quadrilateral are supplementary)

∠NRM + 29˚ + 119˚ = 180˚

∠NRM + 148˚ = 180˚

∠NRM = 180˚ – 148˚

∠NRM = 32˚

**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.

**Solution:**

Construction:- Join AC and BD

∠CAD = 90˚ (angle in a semicircle is a right angle)

∠CBD = 90˚ (angle in a semicircle is a right angle)

Also, AB || CD

∠BAD = ∠ADC = 25˚ (Alternate angles)

∠BAC = ∠BAD + ∠CAD

∠BAC = 25˚ + 90˚

∠BAC = 115˚

∠ADB = 180˚ – 25˚ – ∠BAC

∠ADB = 180˚ – 25˚ – 115˚

∠ADB = 40˚ (Pair of opposite angles in a cyclic quadrilateral are supplementary)

∠AEB = ∠ADB = 40˚ (Angle subtended by the same chord on the circle are equal)

**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.

Solution:

Construction:- Join AC, PQ and BD.

ACQP is a cyclic quadrilateral.

∠CAP + ∠PQC = 180˚_________(1) (Pair of opposite angles in a cyclic quadrilateral are supplementary)

PQDB is a cyclic quadrilateral.

∠PQD + ∠DBP = 180˚_________(2) (Pair of opposite angles in a cyclic quadrilateral are supplementary)

∠PQC + ∠PQD = 180˚_________(3) (CQD is a line)

By Equation (1), (2) and (3)

∠CAP + ∠DBP = 180˚

∠CAB + ∠DBA = 180˚

Hence, we know that, if a transversal line 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.

AC || BD.

**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.

Solution:

Let ABCD be the given cyclic quadrilateral.

PA = PD (given)

∠PAD = ∠PDA ______(1)

∠BAD = 180˚ – ∠PAD

∠CDA = 180˚ – ∠PDA

From equation (1)

∠CDA = 180˚ – ∠PAD (we know that the opposite angles of a cyclic quadrilateral are supplementary)

∠ABC = 180 – ∠CDA

∠ABC = 180 – (180˚ – ∠PAD)

∠ABC = 180 – 180 + ∠PAD

∠ABC = ∠PAD

∠ABC = ∠DCB = ∠PAD = ∠PDA

Hence, AD || BC.

**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.

**Solution: **(i) Angle subtended by the same chord on the circle are equal.

∠PRB = ∠PAB = 35˚

(ii) Angle in a semicircle is a right angle

∠BPA = 90˚

∠BPQ = 90˚

∠PBR = ∠BQP + ∠BPQ

∠PBR = 25˚ + 90˚

∠PBR = 115˚ (Exterior angle of a Δ is equal to the sum of pair of interior opposite angles)

(iii) ∠ABP = 90˚ – ∠BAP

∠ABP = 90˚ – 35˚

∠ABP = 55˚

∠ABP = ∠PBR – ∠ABP

∠ABP = 115˚ – 55˚

∠ABP = 60˚

∠APR = ∠ABR = 60˚ (Angle subtended by the same chord on the circle are equal)

∠BPR = 90˚ – ∠APR

∠BPR = 90˚ – 60˚

∠BPR = 30˚

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

**Solution:**It is given that, PQRS is a cyclic quadrilateral

∠QRS + ∠QPS = 180˚___________ (1) (Pair of opposite angles in a cyclic quadrilateral are supplementary)

∠QPS + ∠SPT = 180˚___________ (2) (QPT is a straight line)

From equation (1) and (2)

∠QRS = ∠SPT__________(iii)

∠RQS = ∠RPS__________(iv) (Angle subtended by the same chord on the circle are equal)

∠RPS = ∠SPT___________(v) (PS is bisector or ∠RPT)

From equation (iii), (iv) and (v)

∠QRS = ∠RQS

Hence, SQ = SR

**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.

**Solutions:**

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

∠DAB + ∠BED = 180˚ (Pair of opposite angles in a cyclic quadrilateral are supplementary)

∠BED = 180˚ – 51˚

∠BED = 129˚

∠CEB = 180˚ – ∠BED

∠CEB = 180˚ – 129˚

∠CEB = 51˚

By angle sum property we know that,

∠OCE = 180˚ – 51˚ – 105˚

∠OCE = 24˚

**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.

**Solution:**∠ACB = 1/2 ∠APB (Angle at the centre is double the angle at the circumference subtended by the same chord)

∠ACB = 1/2 × 150˚

∠ACB = 75˚

∠ACB + ∠BCD = 180˚ (Straight line)

∠BCD = 180˚ – 75˚

∠BCD = 105˚

Also,

∠BCD = 1/2 Reflex ∠BQD

∠BCD = 1/2(360˚-x) (Angle a the centre is double the angle at the circumference subtended by the same chord)

105˚ = 180˚ – x/2

x/2 = 180˚-105˚

x/2 = 75˚

x = 75˚ × 2

x = 150˚

**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.

**Solution:**

(i) It is given that, ∠APB = a˚

obtuse ∠AOB = 2∠APB (Angle at the centre is double the angle at the circumference subtended by the same chord)

obtuse ∠AOB = 2a˚

(ii) OABC is a cyclic quadrilateral

∠AOB + ∠ACB = 180˚ (Pair of opposite angles in a cyclic quadrilateral are supplementary)

∠ACB = 180˚ – 2a˚

(iii) By Construction: Join AB.

∠ADB = ∠ACB

∠ADB = ∠ACB = 180˚ – 2a˚

Angle subtended by the same arc on the circle are equal.

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

**Solution:**∠AOC = 2ABC (Angle at the centre is double the angle at the circumference subtended by the same chord)

∠AOC = 2×55˚

∠AOC = 110˚

x = 110˚

ABCD is a cyclic quadrilateral

∠ADC + ∠ABC = 180˚ (Pair of opposite angles in a cyclic quadrilateral are supplementary)

y = 180˚ – 55˚

y = 125˚

Hence, the value of x is 110˚ and y is 125˚

**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.

**Solution:**

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

∠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

**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.**

**Solution:**(i) ∠DAB = ∠BED = 65˚ (Angles subtended by the same chord on the circle are equal)

(ii) ∠ADB = 90˚ (Angles in a semicircle is a right angle)

∠ADB = 90˚ – ∠DAB

∠ADB = 90˚-65˚

∠ADB = 25˚

AB||DC

∠BDC = ∠ABD = 25˚ (Alternate angles)

**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.

**Solution:**

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

Therefore, ∠EBA = 90˚ – ∠EAB

∠EBA = 90˚-63˚

∠EBA = 27˚

(ii) AB || ED

∠DEB = ∠EBA = 27˚ (Alternate angle)

BCDE is a cyclic quadrilateral

∠DEB + ∠BCD = 180˚ (Pair of opposite angles in a cyclic quadrilateral are supplementary)

∠BCD = 180˚-27˚

Hence, ∠BCD = 153˚

**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.**

**Solution:**

(i) ABCD is a cyclic quadrilateral

∠DCB + ∠DAB = 180˚ (Pair of opposite angles in a cyclic quadrilateral are supplementary)

∠DAB = 180˚ – 120˚

∠DAB = 60˚

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

∠DBA = 90˚ – ∠DAB

∠DBA = 90˚ – 60˚

∠DBA = 30˚

(iii) OD = OB

∠ODB = ∠OBD

∠ABD = 30˚

Also, AB || ED

∠DBC = ∠ODB = 30˚ (By Alternate angles)

(iv) ∠ABD + ∠DBC = 30˚ + 30˚

∠ABD + ∠DBC = 60˚

∠ABC = 60˚

In cyclic quadrilateral ABCD.

∠ADC + ∠ABC = 180˚ (pair of opposite angles in a cyclic quadrilateral are supplementary)

∠ADC = 180˚ – 60˚

∠ADC = 120˚

In ΔAOD,

OA = OD (Radius of the circle)

∠AOD = ∠DAO

∠DAB = 60˚

∠AOD = 60˚

∠ADO = ∠AOD = ∠DAO = 60˚

ΔAOD is an equilateral triangle.

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

**Solution:**

Join IA, IC and CD.

(i) IB is the bisector of ÐABC

**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° – 1/2BAC**

**Solution:**

Construction: – Join PQ and PR

(i) BQ is the bisector of ∠ABC

∠APQ = 1/2 ∠ABC

∠APQ = ∠ABQ (Angle in the same segment)

∠ABC = 2∠APQ

(ii) CR is the bisector of ∠ACB

∠ACR = 1/2 ∠ACB

∠ACR = ∠APR (Angle in the same segment)

∠ACB = 2∠APR

(iii) By Adding equation (i) and (ii)

∠ABC + ∠ACB = 2∠APQ + 2∠APR

∠ABC + ∠ACB = 2(∠APQ + ∠APR)

∠ABC + ∠ACB = 2∠QPR

180˚ – ∠BAC = 2∠QPR

2∠QPR = 180˚ – ∠BAC

**Exercise 17 B**

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

**Solution:**

It is given that, a cyclic trapezium ABCD in which AB || DC and AC and BD are joined.

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 above)

Chord AD = Chord BC

AD = BC

Now in ΔADC and ΔBCD

DC = DC (common)

∠CAD = ∠CBD (angle in the same segment)

AD = BC (proved)

ΔADC≅ΔBCD (SAS congruence rule)

AC = BD (CPCT)

**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.**

**Solution:**

Construction: – Join AE, OB and OC.

(i) Here, AOD is the diameter

∠AED = 90˚ (angle in a semi – circle)

∠DEF = 110˚

∠AEF = ∠DEF – ∠AED

∠AEF = 110˚ – 90˚

∠AEF = 20˚

(ii) Chord AB = Chord BC = Chord CD (given)

∠AOB = ∠BOC = ∠COD (Equal chords subtends equal angle at the centre)

∠AOB + ∠BOC + ∠COD = 180˚ (AOD is a straight line)

∠AOB = ∠BOC = ∠COD = 60˚

In ΔOAB,

OA = OB

Radius of the same circle

∠OAB = ∠OBA

∠OAB + ∠OBA = 180˚ – ∠AOB

∠OAB + ∠OBA = 180˚ – 60˚

∠OAB + ∠OBA = 120˚

∠OAB = ∠OBA = 60˚

In cyclic quadrilateral ADEF,

∠DEF + ∠DAF = 180˚

∠DAF = 180˚ – ∠DEF

∠DAF = 180˚ – 110˚

∠DAF = 70˚

Again,

∠FAB = ∠DAF + ∠OAB

∠FAB = 70˚ + 60˚

∠FAB = 130˚

**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.

Solution:

It is given that, ABCD is a cyclic quadrilateral in which AB||DC. AC and BD are diagonals.

Prove: (i) AD = BC (ii) AC = BD

Proof:-

(i) AB || DC

∠DCA = ∠CAB (alternate angle)

Now, chord AD subtends ∠DCA and chord BC subtends ∠CAB at the circumference of the circle.

∠DCA = ∠CAB

Chord AD = Chord BC

AD = BC

(ii) Now in ΔABC and ΔADB,

AB = AB (Common)

∠ACB = ∠ADB (angles in the same segment)

BC = AD (proved)

ΔACB≅ΔADB

AC = BD (CPCT)

**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.

**Solution:**

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 centre and ∠PTS at the remaining part of the circle.

∠POS = 2∠PTS

∠POS = 2×75˚

∠POS = 105˚

∠POQ + ∠QOR + ∠ROS = 150˚

∠POQ = ∠QOR = ∠ROS = (150˚)/3

∠POQ = ∠QOR = ∠ROS = 50˚

In ΔOPQ,

OP = OQ (radius of a circle)

∠OPQ = ∠OQP

∠OPQ + ∠OQP + ∠POQ = 180˚

∠OPQ + ∠OQP + 50˚ = 180˚

∠OPQ + ∠OQP = 180˚ – 50˚

∠OPQ + ∠OPQ = 130˚

2∠OPQ = 130˚

∠OPQ = (130˚)/2

∠OPQ = 65˚

Similarly we can prove that,

(i) now ∠POS = 150˚

(ii) ∠QOR = 50˚

(ii) ∠POR = ∠PQO + ∠OQR

∠POR = 65˚ + 65˚

∠POR = 130˚

**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.

**Solution:**(i) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∠ACB = 1/2 ∠AOB

Since AB is the side of a regular hexagon,

∠AOB = 60˚

(ii) ∠AOB = 60˚

∠ACB = 1/2 × 60˚

∠ACB = 30˚

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

Solution:

**(i) Arc AB subtends ÐAOB at the centre and ÐACB at the remaining part of the circle.**

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

**Solution:**

In the figure, O is the centre of circle, with AB = BC = CD.

∠ABC = 132˚

(i) In cyclic quadrilateral ABCD,

∠ABC+∠AEC = 180˚ (sum of opposite angles)

132˚+∠AEC = 180˚

∠AEC = 180˚-132˚

∠AEC = 48˚

AB = BC,

∠AEB = ∠BEC (equal chords subtends equal angles)

(ii) Similarly, AB = BC = CD

∠AEB = ∠BED = ∠CED = 24˚

∠AEB = ∠AEB + ∠BEC + ∠CED

∠AEB = 24˚ + 24˚ + 24˚

∠AEB = 72˚

(iii) Arc CD subtends ∠COD at the centre and ∠CED at the remaining part if the circle.

∠COD = 2∠CED

∠COD = 2 × 24˚

∠COD = 48˚

**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.

**Solution:**

∠ACB = 54˚

In cyclic quadrilateral ADBC

∠ADB + ∠ACB = 180˚ (sum of opposite angles)

∠ADB + 54˚ = 180˚

∠ADB = 180˚-54˚

∠ADB = 126˚

**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.

**Solution:**

Join OA, OB and OC

Since AB is the side of a regular pentagon,

∠AOB = 360˚/5

∠AOB = 72˚

AC is the side of a regular hexagon,

∠AOC = 360˚/6

∠AOC = 60˚

∠AOB + ∠AOC + ∠BOC = 360˚

72˚ + 60˚ + ∠BOC = 360˚

132˚ + ∠BOC = 360˚

∠BOC = 360˚ – 132˚

∠BOC = 228˚

Arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.

**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.**

**Solution:**

Join BC, BO, CO and EO

Since BD is the side of a regular hexagon

∠BOD = 360°/6

∠BOD = 60˚

Since DC is the side of a regular pentagon,

∠BOD = (360°)/5

∠BOD = 72˚

In ΔBOD

∠BOD = 60˚ and OB = OD

∠OBD = ∠ODB = 60˚

(i) ΔOCD,∠COD = 72˚ and OC = OD

(iv) In cyclic quadrilateral AECD

∠AEC + ∠ADC = 180˚ (sum of opposite angles)

∠AEC + 54˚ = 180˚

∠AEC = 180˚ – 54˚

∠AEC = 126˚

**Exercise 17 C**

**Question **1. In the given circle with diameter AB, find the value of x.

**Solution:**∠ABD = ∠ACD = 30° (Angle in the same segment)

In ∆ADB,

∠BAD + ∠ADB + ∠DBA = 180° (Angles of a A)

∠ADB = 90° (Angle in a semi-circle)

x + 90° + 30° = 180°

x + 120° = 180°

x = 180° – 120°

x = 60°

Hence, the value of x is 60°.

**Question **2. 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.

**Solution:**

Given that, in the figure ABC is a triangle in which ∠A = 30˚

Need to prove BC is the radius of circumcircle of ΔABC whose centre is O.

Construction: – join OB and OC.

Proof: – ∠BOC = 2∠BAC

∠BOC = 2×30˚

∠BOC = 60˚

Now in ΔOBC,

OB = OC (Radius of the circle)

∠OBC = ∠OCB

In ΔBOC,

∠OBC + ∠OCB + ∠BOC = 180˚ (Angle of a triangle)

∠OBC + ∠OBC + 60˚ = 180˚

2∠OBC = 180˚ – 60˚

2∠OBC = 120˚

∠OBC = (120˚)/2

∠OBC = 60˚

∠OBC = ∠OCB = ∠BOC = 60˚

ΔBOC is an equilateral triangle.

BC = OB = OC

OB and OC are the radius of the circumcircle.

BC is also the radius of the circumcircle.

**Question **3. Prove that the circle drawn on any one a the equal side of an isosceles triangle as diameter bisects the base.

Solution:

It is given that, in ΔABC, AB = AC and a circle with AB as diameter is drawn which intersects the side BC and D.

Need to prove D is the mid – point of BC.

By Construction join AD

Proof: – ∠1 = 90˚

∠1 + ∠2 = 180˚ (linear pair)

∠2 + ∠2 = 180˚

2∠2 = 180˚

∠2 = 180˚/2

∠2 = 90˚

In right ΔABD and ΔACD,

AB = AC (Given)

AD = AD (Common)

ΔABD ≅ ΔACD (by RHS congruence rule)

BD = DC (CPCT)

Hence, D is the mid-point of BC.

**Question **4. In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠ CBE = 65°, calculate ∠DEC.

**Solution:**

By Construction Join OE.

EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.

∠EOC = 2∠EBC

∠EOC = 2 × 65˚

∠EOC = 130˚

In ΔOEC,

OE = OC (Radius of the circle)

∠OEC = ∠OCE

In ΔEOC,

∠OEC + ∠OCE + ∠EOC = 180˚ (angle sum property)

∠OCE + ∠OCE + 130˚ = 180˚

2∠OCE = 180˚ – 130˚

2∠OCE = 50˚

∠OCE = 50˚/2

∠OCE = 25˚

AC || ED (given)

∠DEC is Alternate angles of ∠OCE

∠DEC = ∠OCE

∠DEC = 25˚

**Question **5. The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.

Solution:

It is given that, ABCD is a cyclic quadrilateral and PQRS is a quadrilateral formed by the angle.

By construction bisectors of angle ∠A, ∠B, ∠C and ∠D

Need to prove PQRS is a cyclic quadrilateral.

Proof: –

In ΔAPD,

∠PAD + ∠ADP + ∠APD = 180˚__________(1)

In ΔBQC,

∠QBC + ∠BCQ + ∠BQC = 180˚__________(2)

By adding equation (1) and (2)

∠PAD + ∠ADP + ∠APD + ∠QBC + ∠BCQ + ∠BQC = 180˚ + 180˚

∠PAD + ∠ADP + ∠APD + ∠QBC + ∠BCQ + ∠BQC = 360˚

∠PAD + ∠ADP + ∠QBC + ∠BCQ = 1/2 [∠A + ∠B + ∠C + ∠D]

∠PAD + ∠ADP + ∠QBC + ∠BCQ = 1/2×360˚

∠PAD + ∠ADP + ∠QBC + ∠BCQ = 180˚

∠APD + ∠BQC = 360˚ – 180˚

∠APD + ∠BQC = 180˚

These are the sum of opposite angles of quadrilateral PRQS.

Hence, PQRS is a cyclic quadrilateral.

**Question **6. In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:

(i) ∠BDC

(ii) ∠BEC

(iii) ∠BAC

**Solution:**

(i) It is given that, BD is a diameter of the circle and ∠DBC = 58˚

The angle in a semicircle is a right angle.

∠BCD = 90˚

In ΔBDC,

∠DBC + ∠BCD + ∠BDC = 180˚

58˚ + 90˚ + ∠BDC = 180˚

148˚ + ∠BDC = 180˚

∠BDC = 180˚¬ – 148˚

∠BDC = 32˚

(ii) We know that the opposite angles of a cyclic quadrilateral are supplementary.

In cyclic quadrilateral BECD,

∠BEC + ∠BDC = 180˚

∠BEC + 32˚ = 180˚

∠BEC = 180˚ – 32˚

∠BEC = 148˚

(iii) In cyclic quadrilateral ABEC,

∠BAC + ∠BEC = 180˚

∠BAC + 148˚ = 180˚

∠BAC = 180˚ – 148˚

∠BAC = 32˚

**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 con – cyclic.

Solution:

It is given that, In ΔABC, AB = AC and D and E are points on AB and AC such that AD = AE.

Constriction: – Join DE.

Need to Prove: – B,C,E,D are con – cyclic.

Proof:

In ΔABC, AB = AC

∠B = ∠C (angle opposite to equal sides)

Similarly,

In ΔADE,

AD = AE (given)

∠ADE = ∠AED (angle opposite to equal sides)

In ΔABC,

AP/AB = AE/AC

DE || BC

∠ADE = ∠B (Corresponding angles)

∠B = ∠C (proved above)

Ext. ∠ADE = interior opposite ∠C

BCED is a cyclic quadrilateral.

Hence, B, C, E and D are con – cyclic.

**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.

**Solution**:

In cyclic quadrilateral 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 quadrilateral ABCD,

∠B + ∠D = 180˚

∠B + 92˚ = 180˚

∠B = 180˚ – 92˚

∠B = 88˚

Since AF || CB,∠FAB = ∠B = 88˚

∠FAE = 20˚ (given)

Ext. ∠BAE = ∠BAF + ∠FAE

Ext. ∠BAE = 88˚ + 22˚

Ext. ∠BAE = 108˚

Ext. ∠BAE = ∠BCD

∠BCD = 108˚

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

**Solution:**

By construction join DB and DC, IB and IC,

∠BAC = 66˚,∠ABC = 80˚, I is the in centre of the ΔABC,

(i) Since ∠DBC and ∠DAC are in the same segment,

∠DBC = ∠DAC

(iii) ∠BAC = 66˚ and ∠ABC = 80˚

In ΔABC,

∠ACB = 180˚ – (∠ABC + ∠BAC)

∠ACB = 180˚ – (80˚ + 66˚)

∠ACB = 180˚ – (156˚)

∠ACB = 180˚ – 156˚

∠ACB = 34˚

Since IC bisects the ∠C

∠ICB = 1/2 ∠C

∠ICB = 1/2×34˚

∠ICB = 17˚

Now in ΔIBC,

∠IBC + ∠ICB + ∠BIC = 180˚

40˚ + 17˚ + ∠BIC = 180˚

∠BIC = 180˚ – 57˚

∠BIC = 123˚

**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.**

**Solution:**

It is given that, In the figure, AB = AD = DC = PB and DBC = x˚

Join AC and BD.

Need to find:- the measure of ∠ABD and ∠APB.

Proof: ∠DAC = ∠DBC = x (angles in the same segment)

∠DCA = ∠DAC = x (AD = DC)

We have,

∠ABD = ∠DAC (angles in the same segment)

In ΔABP,

Ext. ∠ABD = ∠BAP + ∠APB

∠BAP = ∠APB (AB = BP)

2×x = ∠APB + ∠APB

2x = 2∠APB

x = ∠APB

∠APB = ∠DBC = x

These are corresponding angles

Hence, AP‖DB┤

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

**Solution:**

It is given that, in the figure ABC, AEQ and CEP are straight line.

Need to prove: ∠APE + ∠CQE = 180˚

By construction: – Join EB

Proof: –

In cyclic quadrilateral ABEP,

∠APE + ∠ABE = 180˚_________(1)

In cyclic quadrilateral BCQE,

∠CQE + ∠CBE = 180˚_________(2)

By adding (1) and (2) equation,

∠APE + ∠ABE + ∠CQE + ∠CBE = 180˚ + 180˚

∠APE + ∠ABE + ∠CBE = 360˚

∠ABE + ∠CBE = 180˚ (linear pair)

∠APE + ∠CQE + 180˚ = 360˚

∠APE + ∠CQE = 360˚ – 180˚

∠APE + ∠CQE = 180˚

Therefore, ∠APE and ∠CQE are supplementary.

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

**Solution:**

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

∠AOC = 2∠ADC

∠AOC = 2 × 32˚

∠AOC = 64˚

Since ∠AOC and ∠BOC are linear pair, we have,

∠AOC + ∠BOC = 180˚

64˚+ ∠BOC = 180˚

∠BOC = 180˚-64˚

∠BOC = 116˚

Hence, the value of ∠BOC is 116˚.

**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.

**Solution:**

PQRS is a cyclic quadrilateral in which ∠PQR = 135˚

Side SP and RQ are produced to meet at A and Side PQ and SR are produced to meet B.

∠A = ∠B = 2:1

Let ∠A = 2x

∠B-x

Now,

In cyclic quadrilateral PQRS,

∠PQR = 135˚

∠S = 180˚-135˚

∠S = 45˚ (Since sum of opposite angle of a cyclic quadrilateral are supplementary)

∠PQR and ∠PQA are linear pair,

∠PQR + ∠PQA = 180˚

135˚ + ∠PQA = 180˚

∠PQA = 180˚-135˚

∠PQA = 45˚

In ΔPBS,

∠P = 180˚-(45˚ + x)

∠P = 180˚-45˚ – x

∠P = 135˚ – x___________(1)

In ΔPQA,

Ext. ∠P = ∠PQA + ∠A

Ext. ∠P = 45˚ + 2x__________(2)

From (1) and (2) we get,

45˚ + 2x = 135˚ – x

2x + x = 135˚ – 45˚

3x = 90˚

x = 90˚/3

x = 30˚

∠A = 2x

∠A = 2×30˚

∠A = 60˚

∠B = x

∠B = 30˚

Hence, the value of ∠A is 60˚ and ∠B is 30˚

**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**

**Solution:**

It is given that, ABCD is a cyclic quadrilateral in which AD || BC bisector of ∠A meets BC at E and the given circle at F.

DF and BF are joined.

Need to prove:- (i) EF = FC (ii) BF = DF

Proof:- ABCD is a cyclic quadrilateral and AD||BC

AF is the bisector of ∠A, ∠BAF = ∠DAF

∠DAE = ∠BAE

∠DAE = ∠AEB (Alternate angles)

(i) In ΔABE,

∠ABE = 180˚ – 2∠AEB

∠CEF = ∠AEB (Vertically opposite angles)

∠ADC = 180˚ – ∠ABC

∠ADC = 180˚ – (180˚ – 2∠AEB)

∠ADC = 180˚ – 180˚ + 2∠AEB

∠ADC = 2∠AEB (Since ADCF is a cyclic quadrilateral)

∠ECF = 180˚ – (∠AFC + ∠CEF)

∠ECF = 180˚ – (180˚ – 2∠AEB + ∠AEB)

∠ECF = 180˚-180˚ + 2∠AEB – ∠AEB)

∠ECF = ∠AEB

EC = EF

(ii) Arc BF = Arc DF (Equal arcs subtends equal angles)

BF = DF (Equal arcs have equal chords)

**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. I f ∠ DCF : ∠F : ∠E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.**Solution:**

It is given that, 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 quadrilateral ABCD,

BC is produced.

∠A = ∠DCF = 3x

In ΔCDF,

Ext. ∠CDA = ∠DCF + ∠F

Ext. ∠CDA = 3x + 5x

Ext. ∠CDA = 8x

In ΔBCE,

Ext. ∠ABC = ∠BCE + ∠E (∠BCE = ∠DCF) (Vertically opposite angles)

Ext. ∠ABC = ∠DCF + ∠E

Ext. ∠ABC = 3x + 4x

Ext. ∠ABC = 7x

In cyclic quadrilateral ABCD,

∠B + ∠D = 180˚

7x + 8x = 180˚

15x = 180˚

x = 180˚/15

x = 12˚

To find ∠A

∠A = 3x

∠A = 3 × 12˚

∠A = 36˚

To find ∠B

∠B = 7x

∠B = 7 × 12˚

∠B = 84˚

To find ∠C

∠C = 180˚ – ∠A

∠C = 180˚ – 36˚

∠C = 144˚

To find ∠D

∠D = 8x

∠D = 8˚ × 12˚

∠D = 96˚

**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 PORS.

**Solution:**

It is given that, In the figure PQRS is a cyclic quadrilateral in which PR is a diameter.

PQ = 7cm

QR = 3RS = 6cm

3RS = 6cm

RS = 2cm

In ΔPQR,

∠Q = 90˚ (Angle in semi circle)

PR^{2} = PQ^{2} + QR^{2} (by Pythagoras theorem)

PR^{2} = (7)^{2} + (6)^{2}

PR^{2} = 49 + 36

PR^{2} = 85

In ΔPSQ,

PR^{2} = PS^{2} + RS^{2} (by Pythagoras theorem)

85 = PS^{2} + (2)^{2}

PS^{2} = 85 – 4

PS^{2} = 81

PS = √81

PS = 9

Perimeter of quadrilateral PQRS = PQ + QR + RS + SP

Perimeter of quadrilateral PQRS = 7 + 9 + 2 + 6

Perimeter of quadrilateral PQRS = 24cm

**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**

**Solution:**

It is given that, in a circle with centre O, AB is the diameter and AC and AD are two chords such that AC = AD.

Need 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

By construction: – Join BC and BD

Proof: – In right angled ΔABC and ΔABD

AC = AD (given)

AB = AB (common)

∠ADC = ∠ADB (each 90˚)

ΔABC≅ΔABD (By RHS congruence rule)

(i) BC = BD (CPCT)

Arc BC = Arc BD (equal chords have equal arcs)

(ii) ∠BAC = ∠BAD

AB is the bisector of ∠CAD

(iii) If Arc AC = 2 arc BC,

∠ABC = 2∠BAC

∠ABC + ∠BAC = 90˚

2∠BAC + ∠BAC = 90˚

3∠BAC = 90˚

∠BAC = 90˚/3

∠BAC = 30˚

**Question **18. In cyclic quadrilateral ABCD; AD = BC, ∠ = 30° and ∠ = 70°; find;

(i) ∠BCD

(ii) ∠BCA

(iii) ∠ABC

(ii) ∠ADC

Solution:

It is given that, ABCD is a cyclic quadrilateral and AD = BC

∠BAC = 30˚

∠CBD = 70˚

We have,

∠DAC = ∠CBD (angles in the same segment)

∠DAC = 70˚ (∠CBD = 70˚)

∠BAD = ∠BAC + ∠DAC

∠BAD = 30˚ + 70˚

∠BAD = 100˚____________(1)

AD = BC, ∠ACD = ∠BDC (equal chords subtends equal angles)

∠ACB = ∠ADB (angles in the same segment)

∠ACD + ∠ACB = ∠BDC + ∠ADB

∠BCD = ∠ADC = 80˚

In ΔBCD,

∠CBD + ∠BCD + ∠BDC = 180˚ (angle sum property)

70˚ + 80˚ + ∠BDC = 180˚

150˚ + ∠BDC = 180˚

∠BDC = 180˚ – 150˚

∠BDC = 30˚

∠ACD = 30˚ (∠ACD = ∠BDC)

∠BCA = ∠BCD – ∠ACD

∠BCA = 80˚ – 30˚

∠BCA = 50˚

∠ADC + ∠ABC = 180˚

80˚ + ∠ABC = 180˚

∠ABC = 180˚ – 80˚

∠ABC = 100˚

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

**Solution:**

It is given that,

∠ACE = 43˚

∠CAF = 62˚

In ΔAEC,

∠ACE + ∠CAE + ∠AEC = 180˚

43˚ + 62˚ + ∠AEC = 180˚

105˚ + ∠AEC = 180˚

∠AEC = 180˚-105˚

∠AEC = 75˚

∠ABD + ∠AED = 180˚

Opposite angles of a cyclic quadrilateral and ∠AED = ∠AEC

a + 75˚ = 180˚

a = 180˚-75˚

a = 105˚

∠EDF = ∠BAE (angles in the alternate segments)

c = 62˚

In ΔBAF,

a + 62˚ + b = 180˚

105˚ + 62˚ + b = 180˚

167˚ + b = 180˚

b = 180˚-167˚

b = 13˚

Hence, the value of a = 105˚, b = 13˚ and c = 62˚

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

**Solution:**

It is given that,

ABCD is a cyclic quadrilateral in which AB||DC. ABCD is an isosceles trapezium

AD = BC

(i) Join BD

We have,

Ext. ∠BCE = ∠BAD (Exterior angle of a cyclic quadrilateral is equal to interior opposite angle)

∠BAD = 80˚

∠BAC = 25˚

∠CAD = ∠BAD – ∠BAC

∠CAD = 80˚ – 25˚

∠CAD = 55˚

(ii) ∠CBD = ∠CAD (angle of the same segment)

∠CBD = 55˚

(iii) ∠ADC = ∠BCD (angles of the isosceles trapezium)

∠ADC = 180˚ – ∠BCE

∠ADC = 180˚ – 80˚

∠ADC = 100˚

**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°.

Solution:

It is given that, ABCD is a cyclic quadrilateral in which AB is the diameter and chord CD is equal to the radius of the circle.

Need to prove ∠APB = 60˚

By construction:- join OC and OD

Proof:- chord CD = CO = DO (radius of the circle)

In ΔDOC, (equilateral triangle)

∠DOC = ∠ODC = ∠DCO = 60˚

Let,

∠A = x and ∠B = y

OA = OB = OC = OD (radius of the same circle)

∠ODA = ∠OAD = x

∠OCB = ∠OBC = y

∠AOD = 180˚-2x

∠BOC = 180˚-2y

AOB is a straight line

∠AOD + ∠BOC + ∠COD = 180˚

180˚-2x + 180˚ – 2y + 60˚ = 180˚

2x + 2y = 240˚

2(x + y) = 240˚

x + y = 240˚/2

x + y = 120˚

∠A + ∠B + ∠P = 180˚ (Angles of a triangle)

120˚ + ∠P = 180˚

∠P = 180˚ – 120˚

∠P = 60˚

Hence, ∠APB is 60˚.

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

**Solution:**

It is given that, in the figure, CP is the bisector of ∠ABC.

Need to prove, DP is the bisector of ∠ADB.

Proof- Since CP is the bisector of ∠ACB

∠ACP = ∠BCP

∠ACP = ∠ADP (angles in the same segment of the circle)

∠BCP = ∠BDP

∠ACP = ∠BCP

∠ADP = ∠BDP

DP is the bisector of ∠ADB.

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

**Solution:**

It is given that, in the figure, ABCD is a cyclic quadrilateral AC and BD are its diagonals.

∠BAC = 30˚

∠CBD = 70˚

Need to Prove: – ∠BCD, ∠BCA, ∠ABC and ∠ADB

∠CAD = ∠CBD = 70˚ (angle in the same segment)

∠BAD = ∠BDC = 30˚

∠BAD = ∠BAC + ∠CAD

∠BAD = 30˚ + 70˚

∠BAD = 100˚

(i) ∠BCD + ∠BAD = 180˚ (opposite angles of cyclic quadrilateral)

∠BCD + ∠BAD = 180˚

∠BCD + 100˚ = 180˚

∠BCD = 180˚ – 100˚

∠BCD = 80˚

(ii) Here, AD = BC, ABCD is an isosceles trapezium and AB||DC

∠BAC = ∠DCA (alternate angles)

∠DCA = 30˚

∠ABC = ∠DCA = 30˚ (angles in the same segment)

∠BCA = ∠BCD – ∠DAC

∠BCA = 80˚ – 30˚

∠BCA = 50˚

(iii) ∠ABC = ∠ABD + ∠CBD

∠ABC = 30˚ + 70˚

∠ABC = 100˚

(iv) ∠ADB = ∠BCA (angle in the same segment)

∠ADB = ∠BCA = 30˚

**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** **Solution:**

(i) AD is parallel to BC

OD is parallel to BC and BD is transversal.

∠ODB = ∠CBD = 32˚ (alternate angles)

In ΔOBD,

OD = OB (Radius of the circle)

∠ODB = ∠OBD = 32˚

(ii) AD is parallel to BC

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 ΔOAB,

OA = OB (Radius of the same circle)

Let us assumed that, ∠OAB = ∠OBA = x

∠OAB + ∠OBA + ∠AOB = 180˚

x + x + 64˚ = 180˚

2x + 64˚ = 180˚

2x = 180˚ – 64˚

2x = 116˚

x = 116˚/2

x = 58˚

∠OAB = 58˚

∠DAB = 58˚

∠DAB = ∠BED = 58˚ (angle inscribed in the same arc are equal)

**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**

**Solution:**

∠DAE and ∠DAB are linear pair

So,

∠DAE + ∠DAB = 180°

∠DAB = 110°

Also,

∠BCD + ∠DAB = 180° (Opposite angles of cyclic quadrilateral BADC)

∠BCD = 70°

∠BCD = 1/2∠BOD (Angles subtended by an arc on the centre and on the circle)

∠BOD = 140°

In ∆BOD,

OB = OD (Radius of same circle)

So,

∠OBD = ∠ODB (isosceles triangle)

∠OBD+∠ODB+∠BOD = 180° (sum of angles of triangle 2∠OBD = 40°)

∠OBD = 20°

Hence, the value of ∠OBD is 20˚.