NCERT Solutions for Class 9 Maths. Chapter 10 Circles. Question Answer.

## NCERT Solutions for Class 9 Maths

### Chapter 10 Circles

**Exercise 10.1**

**Question 1.**

Fill in the blanks.

(i) The centre of a circle lies in ___ of the circle. (exterior/interior)

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in ____ of the circle, (exterior/interior)

(iii) The longest chord of a circle is a ____ of the circle.

(iv) An arc is a ____ when its ends are the ends of a diameter.

(v) Segment of a circle is the region between an arc and ____ of the circle.

(vi) A circle divides the plane, on which it lies, in ____ parts.

Fill in the blanks.

(i) The centre of a circle lies in ___ of the circle. (exterior/interior)

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in ____ of the circle, (exterior/interior)

(iii) The longest chord of a circle is a ____ of the circle.

(iv) An arc is a ____ when its ends are the ends of a diameter.

(v) Segment of a circle is the region between an arc and ____ of the circle.

(vi) A circle divides the plane, on which it lies, in ____ parts.

Solution:

(i) interior

(ii) exterior

(iii) diameter

(iv) semicircle

(v) the chord

(vi) three

(ii) exterior

(iii) diameter

(iv) semicircle

(v) the chord

(vi) three

**Question 2.**

Write True or False. Give reason for your answers.

(i) Line segment joining the centre to any point on the circle is a , radius of the circle.

(ii) A circle has only finite number of equal chords.

(iii) If a circle is divided into three equal arcs, each is a major arc.

(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.

(v) Sector is the region between the chord and its corresponding arc.

(vi) A circle is a plane figure.Write True or False. Give reason for your answers.

(i) Line segment joining the centre to any point on the circle is a , radius of the circle.

(ii) A circle has only finite number of equal chords.

(iii) If a circle is divided into three equal arcs, each is a major arc.

(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.

(v) Sector is the region between the chord and its corresponding arc.

(vi) A circle is a plane figure.

Solution:

(i) True [∵ All points on the circle are equidistant from the centre]

(ii) False [ ∵ A circle can have an infinite number of equal chords]

(iii) False [∵ Each part will be less than a semicircle]

(iv) True [ ∵ Diameter = 2 x Radius]

(v) False [ ∵ The region between the chord and its corresponding arc is a segment]

(vi) True [ ∵ A circle is drawn on a plane]

(ii) False [ ∵ A circle can have an infinite number of equal chords]

(iii) False [∵ Each part will be less than a semicircle]

(iv) True [ ∵ Diameter = 2 x Radius]

(v) False [ ∵ The region between the chord and its corresponding arc is a segment]

(vi) True [ ∵ A circle is drawn on a plane]

**Exercise 10.2**

**Question 1.**

Recall that two circles are congruent, if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres

Recall that two circles are congruent, if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres

Solution:

Given: Two congruent circles with centres O and O’ and radii r, which have chords AB and CD respectively such that AB = CD.

To Prove: ∠AOB = ∠CO’D

Proof: In ∆AOB and ∆CO’D, we have

AB = CD [Given]

OA = O’C [Each equal to r]

OB = O’D [Each equal to r]

∴ ∆AOB ≅ ∆CO’D [By SSS congruence criteria]

⇒ ∠AOB = ∠CO’D [C.P.C.T.]

To Prove: ∠AOB = ∠CO’D

Proof: In ∆AOB and ∆CO’D, we have

AB = CD [Given]

OA = O’C [Each equal to r]

OB = O’D [Each equal to r]

∴ ∆AOB ≅ ∆CO’D [By SSS congruence criteria]

⇒ ∠AOB = ∠CO’D [C.P.C.T.]

**Question 2.**

Prove that, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.Prove that, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Solution:

Given: Two congruent circles with centres O & O’ and radii r which have chords AB and CD respectively such that ∠AOB = ∠CO’D.

To Prove: AB = CD

Proof: In ∆AOB and ∆CO’D, we have

OA = O’C [Each equal to r]

OB = O’D [Each equal to r]

∠AOB = ∠CO’D [Given]

∴ ∆AOB ≅ ∆CO’D [By SAS congruence criteria]

Hence, AB = CD [C.P.C.T.]

To Prove: AB = CD

Proof: In ∆AOB and ∆CO’D, we have

OA = O’C [Each equal to r]

OB = O’D [Each equal to r]

∠AOB = ∠CO’D [Given]

∴ ∆AOB ≅ ∆CO’D [By SAS congruence criteria]

Hence, AB = CD [C.P.C.T.]

**Question 2.**

Suppose you are given a circle. Give a construction to find its centre.Suppose you are given a circle. Give a construction to find its centre.

Solution:

Steps of construction :

Step I : Take any three points on the given circle. Let these points be A, B and C.

Step II : Join AB and BC.

Step III : Draw the perpendicular bisector, PQ of AB.

Step IV: Draw the perpendicular bisector, RS of BC such that it intersects PQ at O.

Thus, ‘O’ is the required centre of the given drcle.

Steps of construction :

Step I : Take any three points on the given circle. Let these points be A, B and C.

Step II : Join AB and BC.

Step III : Draw the perpendicular bisector, PQ of AB.

Step IV: Draw the perpendicular bisector, RS of BC such that it intersects PQ at O.

Thus, ‘O’ is the required centre of the given drcle.

**Question 3.**

If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Solution:

We have two circles with centres O and O’, intersecting at A and B.

∴ AB is the common chord of two circles and OO’ is the line segment joining their centres.

Let OO’ and AB intersect each other at M.

∴ To prove that OO’ is the perpendicular bisector of AB,

we join OA, OB, O’A and O’B. Now, in ∆QAO’ and ∆OBO’,

we have

OA = OB [Radii of the same circle]

O’A = O’B [Radii of the same circle]

OO’ = OO’ [Common]

∴ ∆OAO’ ≅ ∆OBO’ [By SSS congruence criteria]

⇒ ∠1 = ∠2 , [C.P.C.T.]

Now, in ∆AOM and ∆BOM, we have

OA = OB [Radii of the same circle]

OM = OM [Common]

∠1 = ∠2 [Proved above]

∴ ∆AOM = ∆BOM [By SAS congruence criteria]

⇒ ∠3 = ∠4 [C.P.C.T.]

But ∠3 + ∠4 = 180° [Linear pair]

∴∠3=∠4 = 90°

⇒ AM ⊥ OO’

Also, AM = BM [C.P.C.T.]

⇒ M is the mid-point of AB.

Thus, OO’ is the perpendicular bisector of AB.

We have two circles with centres O and O’, intersecting at A and B.

∴ AB is the common chord of two circles and OO’ is the line segment joining their centres.

Let OO’ and AB intersect each other at M.

∴ To prove that OO’ is the perpendicular bisector of AB,

we join OA, OB, O’A and O’B. Now, in ∆QAO’ and ∆OBO’,

we have

OA = OB [Radii of the same circle]

O’A = O’B [Radii of the same circle]

OO’ = OO’ [Common]

∴ ∆OAO’ ≅ ∆OBO’ [By SSS congruence criteria]

⇒ ∠1 = ∠2 , [C.P.C.T.]

Now, in ∆AOM and ∆BOM, we have

OA = OB [Radii of the same circle]

OM = OM [Common]

∠1 = ∠2 [Proved above]

∴ ∆AOM = ∆BOM [By SAS congruence criteria]

⇒ ∠3 = ∠4 [C.P.C.T.]

But ∠3 + ∠4 = 180° [Linear pair]

∴∠3=∠4 = 90°

⇒ AM ⊥ OO’

Also, AM = BM [C.P.C.T.]

⇒ M is the mid-point of AB.

Thus, OO’ is the perpendicular bisector of AB.

**Exercise 10.4**

**Question 1.**

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Solution:

We have two intersecting circles with centres at O and O’ respectively. Let PQ be the common chord.

∵ In two intersecting circles, the line joining their centres is perpendicular bisector of the common chord.

∴∠OLP = ∠OLQ = 90° and PL = LQ

Now, in right ∆OLP, we have

PL2 + OL2 = 2

⇒ PL2 + (4 – x)2 = 52

⇒ PL2 = 52 – (4 – x)2

⇒ PL2 = 25 -16 – x2 + 8x

⇒ PL2 = 9 – x2 + 8x …(i)

Again, in right ∆O’LP,

PL2 = PO‘2 – LO‘2

= 32 – x2 = 9 – x2 …(ii)

From (i) and (ii), we have

9 – x2 + 8x = 9 – x2

⇒ 8x = 0

⇒ x = 0

⇒ L and O’ coincide.

∴ PQ is a diameter of the smaller circle.

⇒ PL = 3 cm

But PL = LQ

∴ LQ = 3 cm

∴ PQ = PL + LQ = 3cm + 3cm = 6cm

Thus, the required length of the common chord = 6 cm.

We have two intersecting circles with centres at O and O’ respectively. Let PQ be the common chord.

∵ In two intersecting circles, the line joining their centres is perpendicular bisector of the common chord.

∴∠OLP = ∠OLQ = 90° and PL = LQ

Now, in right ∆OLP, we have

PL2 + OL2 = 2

⇒ PL2 + (4 – x)2 = 52

⇒ PL2 = 52 – (4 – x)2

⇒ PL2 = 25 -16 – x2 + 8x

⇒ PL2 = 9 – x2 + 8x …(i)

Again, in right ∆O’LP,

PL2 = PO‘2 – LO‘2

= 32 – x2 = 9 – x2 …(ii)

From (i) and (ii), we have

9 – x2 + 8x = 9 – x2

⇒ 8x = 0

⇒ x = 0

⇒ L and O’ coincide.

∴ PQ is a diameter of the smaller circle.

⇒ PL = 3 cm

But PL = LQ

∴ LQ = 3 cm

∴ PQ = PL + LQ = 3cm + 3cm = 6cm

Thus, the required length of the common chord = 6 cm.

**Question 2.**

If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Solution:

Given: A circle with centre O and equal chords AB and CD intersect at E.

To Prove: AE = DE and CE = BE

Construction : Draw OM ⊥ AB and ON ⊥ CD.

Join OE.

Proof: Since AB = CD [Given]

∴ OM = ON [Equal chords are equidistant from the centre]

Now, in ∆OME and ∆ONE, we have

∠OME = ∠ONE [Each equal to 90°]

OM = ON [Proved above]

OE = OE [Common hypotenuse]

∴ ∆OME ≅ ∆ONE [By RHS congruence criteria]

⇒ ME = NE [C.P.C.T.]

Adding AM on both sides, we get

⇒ AM + ME = AM + NE

⇒ AE = DN + NE = DE

∵ AB = CD ⇒ =

⇒ AM = DN

⇒ AE = DE …(i)

Now, AB – AE = CD – DE

⇒ BE = CE …….(ii)

From (i) and (ii), we have

AE = DE and CE = BE

Given: A circle with centre O and equal chords AB and CD intersect at E.

To Prove: AE = DE and CE = BE

Construction : Draw OM ⊥ AB and ON ⊥ CD.

Join OE.

Proof: Since AB = CD [Given]

∴ OM = ON [Equal chords are equidistant from the centre]

Now, in ∆OME and ∆ONE, we have

∠OME = ∠ONE [Each equal to 90°]

OM = ON [Proved above]

OE = OE [Common hypotenuse]

∴ ∆OME ≅ ∆ONE [By RHS congruence criteria]

⇒ ME = NE [C.P.C.T.]

Adding AM on both sides, we get

⇒ AM + ME = AM + NE

⇒ AE = DN + NE = DE

∵ AB = CD ⇒ =

⇒ AM = DN

⇒ AE = DE …(i)

Now, AB – AE = CD – DE

⇒ BE = CE …….(ii)

From (i) and (ii), we have

AE = DE and CE = BE

**Question 3.**

If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Solution:

Given: A circle with centre O and equal chords AB and CD are intersecting at E.

To Prove : ∠OEA = ∠OED

Construction: Draw OM ⊥ AB and ON ⊥ CD.

Join OE.

Proof: In ∆OME and ∆ONE,

OM = ON

[Equal chords are equidistant from the centre]

OE = OE [Common hypotenuse]

∠OME = ∠ONE [Each equal to 90°]

∴ ∆OME ≅ ∆ONE [By RHS congruence criteria]

⇒ ∠OEM = ∠OEN [C.P.C.T.]

⇒ ∠OEA = ∠OED

Given: A circle with centre O and equal chords AB and CD are intersecting at E.

To Prove : ∠OEA = ∠OED

Construction: Draw OM ⊥ AB and ON ⊥ CD.

Join OE.

Proof: In ∆OME and ∆ONE,

OM = ON

[Equal chords are equidistant from the centre]

OE = OE [Common hypotenuse]

∠OME = ∠ONE [Each equal to 90°]

∴ ∆OME ≅ ∆ONE [By RHS congruence criteria]

⇒ ∠OEM = ∠OEN [C.P.C.T.]

⇒ ∠OEA = ∠OED

**Question 4.**

If a line intersects two concentric circles (circles with the same centre) with centre 0 at A, B, C and D, prove that AB = CD (see figure).If a line intersects two concentric circles (circles with the same centre) with centre 0 at A, B, C and D, prove that AB = CD (see figure).

Solution:

Given : Two circles with the common centre O.

A line D intersects the outer circle at A and D and the inner circle at B and C.

To Prove : AB = CD.

Construction:

Draw OM ⊥ l.

Proof: For the outer circle,

OM ⊥ l [By construction]

∴ AM = MD …(i)

[Perpendicular from the centre to the chord bisects the chord]

For the inner circle,

OM ⊥ l [By construction]

∴ BM = MC …(ii)

[Perpendicular from the centre to the chord bisects the chord]

Subtracting (ii) from (i), we have

AM – BM = MD – MC

⇒ AB = CD

Given : Two circles with the common centre O.

A line D intersects the outer circle at A and D and the inner circle at B and C.

To Prove : AB = CD.

Construction:

Draw OM ⊥ l.

Proof: For the outer circle,

OM ⊥ l [By construction]

∴ AM = MD …(i)

[Perpendicular from the centre to the chord bisects the chord]

For the inner circle,

OM ⊥ l [By construction]

∴ BM = MC …(ii)

[Perpendicular from the centre to the chord bisects the chord]

Subtracting (ii) from (i), we have

AM – BM = MD – MC

⇒ AB = CD

**Question 5.**

Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?

Solution:

Let the three girls Reshma, Salma and Mandip be positioned at R, S and M respectively on the circle with centre O and radius 5 m such that

RS = SM = 6 m [Given]

Equal chords of a circle subtend equal angles at the centre.

∴ ∠1 = ∠2

In ∆POR and ∆POM, we have

OP = OP [Common]

OR = OM [Radii of the same circle]

∠1 = ∠2 [Proved above]

∴ ∆POR ≅ ∆POM [By SAS congruence criteria]

∴ PR = PM and

∠OPR = ∠OPM [C.P.C.T.]

∵∠OPR + ∠OPM = 180° [Linear pair]

∴∠OPR = ∠OPM = 90°

⇒ OP ⊥ RM

Now, in ∆RSP and ∆MSP, we have

RS = MS [Each 6 cm]

SP = SP [Common]

PR = PM [Proved above]

∴ ∆RSP ≅ ∆MSP [By SSS congruence criteria]

⇒ ∠RPS = ∠MPS [C.P.C.T.]

But ∠RPS + ∠MPS = 180° [Linear pair]

⇒ ∠RPS = ∠MPS = 90°

SP passes through O.

Let OP = x m

∴ SP = (5 – x)m

Now, in right ∆OPR, we have

x2 + RP2 = 52

RP2 = 52 – x2

In right ∆SPR, we have

(5 – x)2 + RP2 = 62

⇒ RP2 = 62 – (5 – x)2 ……..(ii)

From (i) and (ii), we get

⇒ 52 – x2 = 62 – (5 – x)2

⇒ 25 – x2 = 36 – [25 – 10x + x2]

⇒ – 10x + 14 = 0

⇒ 10x = 14 ⇒ x = = 1.4

Now, RP2 = 52 – x2

⇒ RP2 = 25 – (1.4)2

⇒ RP2 = 25 – 1.96 = 23.04

∴ RP = = 4.8

∴ RM = 2RP = 2 x 4.8 = 9.6

Thus, distance between Reshma and Mandip is 9.6 m.

RS = SM = 6 m [Given]

Equal chords of a circle subtend equal angles at the centre.

∴ ∠1 = ∠2

In ∆POR and ∆POM, we have

OP = OP [Common]

OR = OM [Radii of the same circle]

∠1 = ∠2 [Proved above]

∴ ∆POR ≅ ∆POM [By SAS congruence criteria]

∴ PR = PM and

∠OPR = ∠OPM [C.P.C.T.]

∵∠OPR + ∠OPM = 180° [Linear pair]

∴∠OPR = ∠OPM = 90°

⇒ OP ⊥ RM

Now, in ∆RSP and ∆MSP, we have

RS = MS [Each 6 cm]

SP = SP [Common]

PR = PM [Proved above]

∴ ∆RSP ≅ ∆MSP [By SSS congruence criteria]

⇒ ∠RPS = ∠MPS [C.P.C.T.]

But ∠RPS + ∠MPS = 180° [Linear pair]

⇒ ∠RPS = ∠MPS = 90°

SP passes through O.

Let OP = x m

∴ SP = (5 – x)m

Now, in right ∆OPR, we have

x2 + RP2 = 52

RP2 = 52 – x2

In right ∆SPR, we have

(5 – x)2 + RP2 = 62

⇒ RP2 = 62 – (5 – x)2 ……..(ii)

From (i) and (ii), we get

⇒ 52 – x2 = 62 – (5 – x)2

⇒ 25 – x2 = 36 – [25 – 10x + x2]

⇒ – 10x + 14 = 0

⇒ 10x = 14 ⇒ x = = 1.4

Now, RP2 = 52 – x2

⇒ RP2 = 25 – (1.4)2

⇒ RP2 = 25 – 1.96 = 23.04

∴ RP = = 4.8

∴ RM = 2RP = 2 x 4.8 = 9.6

Thus, distance between Reshma and Mandip is 9.6 m.

**Question 6.**

A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Solution:

Let Ankur, Syed and David are sitting at A, S and D respectively in the circular park with centre O such that AS = SD = DA

i. e., ∆ASD is an equilateral triangle.

Let the length of each side of the equilateral triangle be 2x.

Draw AM ⊥ SD.

Since ∆ASD is an equilateral triangle.

∴ AM passes through O.

⇒ SM = SD = (2x)

⇒ SM = x

Now, in right ∆ASM, we have

AM2 + SM2 = AS2 [Using Pythagoras theorem]

⇒ AM2= AS2 – SM2= (2x)2 – x2

= 4x2 – x2 = 3x2

⇒ AM = m

Now, OM = AM – OA= ( – 20)m

Again, in right ∆OSM, we have

OS2 = SM2 + OM2 [using Pythagoras theorem]

202 = x2 + ( – 20)2

⇒ 400 = x2 + 3x2 – 40 + 400

⇒ 4x2 = 40

⇒ x = 10√3 m

Now, SD = 2x = 2 x 10√3 m = 20√3 m

Thus, the length of the string of each phone = 20√3 m

Let Ankur, Syed and David are sitting at A, S and D respectively in the circular park with centre O such that AS = SD = DA

i. e., ∆ASD is an equilateral triangle.

Let the length of each side of the equilateral triangle be 2x.

Draw AM ⊥ SD.

Since ∆ASD is an equilateral triangle.

∴ AM passes through O.

⇒ SM = SD = (2x)

⇒ SM = x

Now, in right ∆ASM, we have

AM2 + SM2 = AS2 [Using Pythagoras theorem]

⇒ AM2= AS2 – SM2= (2x)2 – x2

= 4x2 – x2 = 3x2

⇒ AM = m

Now, OM = AM – OA= ( – 20)m

Again, in right ∆OSM, we have

OS2 = SM2 + OM2 [using Pythagoras theorem]

202 = x2 + ( – 20)2

⇒ 400 = x2 + 3x2 – 40 + 400

⇒ 4x2 = 40

⇒ x = 10√3 m

Now, SD = 2x = 2 x 10√3 m = 20√3 m

Thus, the length of the string of each phone = 20√3 m

**Exercise 10.5**

**Question 1.**

In figure A,B and C are three points on a circle with centre 0 such that ∠BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ ADC.

In figure A,B and C are three points on a circle with centre 0 such that ∠BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ ADC.

Solution:

We have a circle with centre O, such that

∠AOB = 60° and ∠BOC = 30°

∵∠AOB + ∠BOC = ∠AOC

∴ ∠AOC = 60° + 30° = 90°

The angle subtended by an arc at the circle is half the angle subtended by it at the centre.

∴ ∠ ADC = (∠AOC) = (90°) = 45°

∠AOB = 60° and ∠BOC = 30°

∵∠AOB + ∠BOC = ∠AOC

∴ ∠AOC = 60° + 30° = 90°

The angle subtended by an arc at the circle is half the angle subtended by it at the centre.

∴ ∠ ADC = (∠AOC) = (90°) = 45°

**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.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.

Solution:

We have a circle having a chord AB equal to radius of the circle.

∴ AO = BO = AB

⇒ ∆AOB is an equilateral triangle.

Since, each angle of an equilateral triangle is 60°.

⇒ ∠AOB = 60°

Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ACB at a point on the minor arc of the circle.

Hence, the angle subtended by the chord on the minor arc = 150°.

Similarly, ∠ADB = [∠AOB] = x 60° = 30°

Hence, the angle subtended by the chord on the major arc = 30°

We have a circle having a chord AB equal to radius of the circle.

∴ AO = BO = AB

⇒ ∆AOB is an equilateral triangle.

Since, each angle of an equilateral triangle is 60°.

⇒ ∠AOB = 60°

Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ACB at a point on the minor arc of the circle.

Hence, the angle subtended by the chord on the minor arc = 150°.

Similarly, ∠ADB = [∠AOB] = x 60° = 30°

Hence, the angle subtended by the chord on the major arc = 30°

**Question 3.**

In figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.In figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Solution:

The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point pn the circumference.

∴ reflex ∠POR = 2∠PQR

But ∠PQR = 100°

∴ reflex ∠POR = 2 x 100° = 200°

Since, ∠POR + reflex ∠POR = 360°

⇒ ∠POR = 360° – 200°

⇒ ∠POR = 160°

Since, OP = OR [Radii of the same circle]

∴ In ∆POR, ∠OPR = ∠ORP

[Angles opposite to equal sides of a triangle are equal]

Also, ∠OPR + ∠ORP + ∠POR = 180°

[Sum of the angles of a triangle is 180°]

⇒ ∠OPR + ∠ORP + 160° = 180°

⇒ 2∠OPR = 180° -160° = 20° [∠OPR = ∠ORP]

⇒ ∠OPR = = 10°

The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point pn the circumference.

∴ reflex ∠POR = 2∠PQR

But ∠PQR = 100°

∴ reflex ∠POR = 2 x 100° = 200°

Since, ∠POR + reflex ∠POR = 360°

⇒ ∠POR = 360° – 200°

⇒ ∠POR = 160°

Since, OP = OR [Radii of the same circle]

∴ In ∆POR, ∠OPR = ∠ORP

[Angles opposite to equal sides of a triangle are equal]

Also, ∠OPR + ∠ORP + ∠POR = 180°

[Sum of the angles of a triangle is 180°]

⇒ ∠OPR + ∠ORP + 160° = 180°

⇒ 2∠OPR = 180° -160° = 20° [∠OPR = ∠ORP]

⇒ ∠OPR = = 10°

**Question 4.**

In figure, ∠ABC = 69°,∠ACB = 31°, find ∠BDC.In figure, ∠ABC = 69°,∠ACB = 31°, find ∠BDC.

Solution:

In ∆ABC, ∠ABC + ∠ACB + ∠BAC = 180°

⇒ 69° + 31° + ∠BAC = 180°

⇒ ∠BAC = 180° – 100° = 80°

Since, angles in the same segment are equal.

∴∠BDC = ∠BAC ⇒ ∠BDC = 80°

In ∆ABC, ∠ABC + ∠ACB + ∠BAC = 180°

⇒ 69° + 31° + ∠BAC = 180°

⇒ ∠BAC = 180° – 100° = 80°

Since, angles in the same segment are equal.

∴∠BDC = ∠BAC ⇒ ∠BDC = 80°

**Question 5.**

In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠BAC.In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠BAC.

Solution:

∠BEC = ∠EDC + ∠ECD

[Sum of interior opposite angles is equal to exterior angle]

⇒ 130° = ∠EDC + 20°

⇒ ∠EDC = 130° – 20° = 110°

⇒ ∠BDC = 110°

Since, angles in the same segment are equal.

∴ ∠BAC = ∠BDC

⇒ ∠BAC = 110°

∠BEC = ∠EDC + ∠ECD

[Sum of interior opposite angles is equal to exterior angle]

⇒ 130° = ∠EDC + 20°

⇒ ∠EDC = 130° – 20° = 110°

⇒ ∠BDC = 110°

Since, angles in the same segment are equal.

∴ ∠BAC = ∠BDC

⇒ ∠BAC = 110°

**Question 6.**

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.

Solution:

Since angles in the same segment of a circle are equal.

∴ ∠BAC = ∠BDC

⇒ ∠BDC = 30°

lso, ∠DBC = 70° [Given]

In ∆BCD, we have

∠BCD + ∠DBC + ∠CDB = 180° [Sum of angles of a triangle is 180°]

⇒ ∠BCD + 70° + 30° = 180°

⇒ ∠BCD = 180° -100° = 80°

Now, in ∆ABC,

AB = BC [Given]

∴ ∠BCA = ∠BAC [Angles opposite to equal sides of a triangle are equal]

⇒ ∠BCA = 30° [∵ ∠B AC = 30°]

Now, ∠BCA + ∠BCD = ∠BCD

⇒ 30° + ∠ECD = 80°

⇒ ∠BCD = 80° – 30° = 50°

∴ ∠BAC = ∠BDC

⇒ ∠BDC = 30°

lso, ∠DBC = 70° [Given]

In ∆BCD, we have

∠BCD + ∠DBC + ∠CDB = 180° [Sum of angles of a triangle is 180°]

⇒ ∠BCD + 70° + 30° = 180°

⇒ ∠BCD = 180° -100° = 80°

Now, in ∆ABC,

AB = BC [Given]

∴ ∠BCA = ∠BAC [Angles opposite to equal sides of a triangle are equal]

⇒ ∠BCA = 30° [∵ ∠B AC = 30°]

Now, ∠BCA + ∠BCD = ∠BCD

⇒ 30° + ∠ECD = 80°

⇒ ∠BCD = 80° – 30° = 50°

**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.If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Solution:

Since AC and BD are diameters.

⇒ AC = BD …(i) [All diameters of a circle are equal]

Also, ∠BAD = 90° [Angle formed in a semicircle is 90°]

Similarly, ∠ABC = 90°, ∠BCD = 90°

and ∠CDA = 90°

Now, in ∆ABC and ∆BAD, we have

AC = BD [From (i)]

AB = BA [Common hypotenuse]

∠ABC = ∠BAD [Each equal to 90°]

∴ ∆ABC ≅ ∆BAD [By RHS congruence criteria]

⇒ BC = AD [C.P.C.T.]

Similarly, AB = DC

Thus, the cyclic quadrilateral ABCD is such that its opposite sides are equal and each of its angle is a right angle.

∴ ABCD is a rectangle.

⇒ AC = BD …(i) [All diameters of a circle are equal]

Also, ∠BAD = 90° [Angle formed in a semicircle is 90°]

Similarly, ∠ABC = 90°, ∠BCD = 90°

and ∠CDA = 90°

Now, in ∆ABC and ∆BAD, we have

AC = BD [From (i)]

AB = BA [Common hypotenuse]

∠ABC = ∠BAD [Each equal to 90°]

∴ ∆ABC ≅ ∆BAD [By RHS congruence criteria]

⇒ BC = AD [C.P.C.T.]

Similarly, AB = DC

Thus, the cyclic quadrilateral ABCD is such that its opposite sides are equal and each of its angle is a right angle.

∴ ABCD is a rectangle.

**Question 8.**

If the non – parallel sides of a trapezium are equal, prove that it is cyclic.If the non – parallel sides of a trapezium are equal, prove that it is cyclic.

Solution:

We have a trapezium ABCD such that AB ॥ CD and AD = BC.

Let us draw BE ॥ AD such that ABED is a parallelogram.

∵ The opposite angles and opposite sides of a parallelogram are equal.

∴ ∠BAD = ∠BED …(i)

and AD = BE …(ii)

But AD = BC [Given] …(iii)

∴ From (ii) and (iii), we have BE = BC

⇒ ∠BCE = ∠BEC … (iv) [Angles opposite to equal sides of a triangle are equal]

Now, ∠BED + ∠BEC = 180° [Linear pair]

⇒ ∠BAD + ∠BCE = 180° [Using (i) and (iv)]

i.e., A pair of opposite angles of a quadrilateral ABCD is 180°.

∴ ABCD is cyclic.

⇒ The trapezium ABCD is cyclic.

We have a trapezium ABCD such that AB ॥ CD and AD = BC.

Let us draw BE ॥ AD such that ABED is a parallelogram.

∵ The opposite angles and opposite sides of a parallelogram are equal.

∴ ∠BAD = ∠BED …(i)

and AD = BE …(ii)

But AD = BC [Given] …(iii)

∴ From (ii) and (iii), we have BE = BC

⇒ ∠BCE = ∠BEC … (iv) [Angles opposite to equal sides of a triangle are equal]

Now, ∠BED + ∠BEC = 180° [Linear pair]

⇒ ∠BAD + ∠BCE = 180° [Using (i) and (iv)]

i.e., A pair of opposite angles of a quadrilateral ABCD is 180°.

∴ ABCD is cyclic.

⇒ The trapezium ABCD is cyclic.

**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 ∠ ACP = ∠QCD.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 ∠ ACP = ∠QCD.

Solution:

Since, angles in the same segment of a circle are equal.

∴ ∠ACP = ∠ABP …(i)

Similarly, ∠QCD = ∠QBD …(ii)

Since, ∠ABP = ∠QBD …(iii) [Vertically opposite angles]

∴ From (i), (ii) and (iii), we have

∠ACP = ∠QCD

Since, angles in the same segment of a circle are equal.

∴ ∠ACP = ∠ABP …(i)

Similarly, ∠QCD = ∠QBD …(ii)

Since, ∠ABP = ∠QBD …(iii) [Vertically opposite angles]

∴ From (i), (ii) and (iii), we have

∠ACP = ∠QCD

**Question 10.**

If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Solution:

We have ∆ABC, and two circles described with diameter as AB and AC respectively. They intersect at a point D, other than A.

Let us join A and D.

∵ AB is a diameter.

∴∠ADB is an angle formed in a semicircle.

⇒ ∠ADB = 90° ……(i)

Similarly, ∠ADC = 90° ….(ii)

Adding (i) and (ii), we have

∠ADB + ∠ADC = 90° + 90° = 180°

i. e., B, D and C are collinear points.

⇒ BC is a straight line. Thus, D lies on BC.

We have ∆ABC, and two circles described with diameter as AB and AC respectively. They intersect at a point D, other than A.

Let us join A and D.

∵ AB is a diameter.

∴∠ADB is an angle formed in a semicircle.

⇒ ∠ADB = 90° ……(i)

Similarly, ∠ADC = 90° ….(ii)

Adding (i) and (ii), we have

∠ADB + ∠ADC = 90° + 90° = 180°

i. e., B, D and C are collinear points.

⇒ BC is a straight line. Thus, D lies on BC.

**Question 11.**

ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

Solution:

We have ∆ABC and ∆ADC such that they are having AC as their common hypotenuse and ∠ADC = 90° = ∠ABC

∴ Both the triangles are in semi-circle.

Case – I: If both the triangles are in the same semi-circle.

⇒ A, B, C and D are concyclic.

Join BD.

DC is a chord.

∴ ∠CAD and ∠CBD are formed in the same segment.

⇒ ∠CAD = ∠CBD

Case – II : If both the triangles are not in the same semi-circle.

⇒ A,B,C and D are concyclic. Join BD. DC is a chord.

∴ ∠CAD and ∠CBD are formed in the same segment.

⇒ ∠CAD = ∠CBD

We have ∆ABC and ∆ADC such that they are having AC as their common hypotenuse and ∠ADC = 90° = ∠ABC

∴ Both the triangles are in semi-circle.

Case – I: If both the triangles are in the same semi-circle.

⇒ A, B, C and D are concyclic.

Join BD.

DC is a chord.

∴ ∠CAD and ∠CBD are formed in the same segment.

⇒ ∠CAD = ∠CBD

Case – II : If both the triangles are not in the same semi-circle.

⇒ A,B,C and D are concyclic. Join BD. DC is a chord.

∴ ∠CAD and ∠CBD are formed in the same segment.

⇒ ∠CAD = ∠CBD

**Question 12.**

Prove that a cyclic parallelogram is a rectangle.Prove that a cyclic parallelogram is a rectangle.

Solution:

We have a cyclic parallelogram ABCD. Since, ABCD is a cyclic quadrilateral.

∴ Sum of its opposite angles is 180°.

⇒ ∠A + ∠C = 180° …(i)

But ∠A = ∠C …(ii)

[Opposite angles of a parallelogram are equal]

From (i) and (ii), we have

∠A = ∠C = 90°

Similarly,

∠B = ∠D = 90°

⇒ Each angle of the parallelogram ABCD is 90°.

Thus, ABCD is a rectangle.

We have a cyclic parallelogram ABCD. Since, ABCD is a cyclic quadrilateral.

∴ Sum of its opposite angles is 180°.

⇒ ∠A + ∠C = 180° …(i)

But ∠A = ∠C …(ii)

[Opposite angles of a parallelogram are equal]

From (i) and (ii), we have

∠A = ∠C = 90°

Similarly,

∠B = ∠D = 90°

⇒ Each angle of the parallelogram ABCD is 90°.

Thus, ABCD is a rectangle.

**Exercise 10.6****Question 1.**

Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.Solution:

Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Given : Two circles with centres O and O’ respectively such that they intersect each other at P and Q.

To Prove: ∠OPO’ = ∠OQO’.

Construction : Join OP, O’P, OQ, O’Q and OO’.

Proof: In ∆OPO’ and ∆OQO’, we have

OP = OQ [Radii of the same circle]

O’P = O’Q [Radii of the same circle]

OO’ = OO’ [Common]

∴ AOPO’ = AOQO’ [By SSS congruence criteria]

⇒ ∠OPO’ = ∠OQO’ [C.P.C.T.]

**Question 2.**

Two chords AB and CD of lengths 5 cm and 11 cm, respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Two chords AB and CD of lengths 5 cm and 11 cm, respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Solution:

We have a circle with centre O.

AB || CD and the perpendicular distance between AB and CD is 6 cm and AB = 5 cm, CD = 11 cm.

Let r cm be the radius of the circle.

Let us draw OP ⊥ AB and OQ ⊥ CD such that

PQ = 6 cm

Join OA and OC.

Let OQ = x cm

∴ OP = (6 – x) cm

∵ The perpendicular drawn from the centre of a circle to chord bisects the chord.

We have a circle with centre O.

AB || CD and the perpendicular distance between AB and CD is 6 cm and AB = 5 cm, CD = 11 cm.

Let r cm be the radius of the circle.

Let us draw OP ⊥ AB and OQ ⊥ CD such that

PQ = 6 cm

Join OA and OC.

Let OQ = x cm

∴ OP = (6 – x) cm

∵ The perpendicular drawn from the centre of a circle to chord bisects the chord.

**Question 3.**

The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre ?The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre ?

Solution:

We have a circle with centre O. Parallel chords AB and CD are such that the smaller chord is 4 cm away from the centre.

Let r cm be the radius of the circle and draw OP ⊥ AB and join OA and OC.

∵ OP ⊥ AB

∴ P is the mid-point of AB.

⇒ AP = = = 3 cm

Similarly, CQ = = = 4 cm

Now in ∆OPA, we have OA2 = OP2 + AP2

⇒ r2 = 42 + 32

⇒ r2 = 16 + 9 = 25

⇒ r = =5

Again, in ∆CQO, we have OC2 = OQ2 + CQ2

⇒ r2 = OQ2 + 42

⇒ OQ2 = r2 – 42

⇒ OQ2 = 52 – 42 = 25 – 16 = 9 [∵ r = 5]

⇒ OQ

⇒ √9 = 3

The distance of the other chord (CD) from the centre is 3 cm.

Note: In case if we take the two parallel chords on either side of the centre, then

In ∆POA, OA2 = OP2 + PA2

⇒ r2 = 42 + 32 = 52

⇒ r = 5

In ∆QOC, OC2 = CQ2 + OQ2

⇒ OQ2 = 42 + OQ2

⇒ OQ2 = 52 – 42 = 9

⇒ OQ = 3

We have a circle with centre O. Parallel chords AB and CD are such that the smaller chord is 4 cm away from the centre.

Let r cm be the radius of the circle and draw OP ⊥ AB and join OA and OC.

∵ OP ⊥ AB

∴ P is the mid-point of AB.

⇒ AP = = = 3 cm

Similarly, CQ = = = 4 cm

Now in ∆OPA, we have OA2 = OP2 + AP2

⇒ r2 = 42 + 32

⇒ r2 = 16 + 9 = 25

⇒ r = =5

Again, in ∆CQO, we have OC2 = OQ2 + CQ2

⇒ r2 = OQ2 + 42

⇒ OQ2 = r2 – 42

⇒ OQ2 = 52 – 42 = 25 – 16 = 9 [∵ r = 5]

⇒ OQ

⇒ √9 = 3

The distance of the other chord (CD) from the centre is 3 cm.

Note: In case if we take the two parallel chords on either side of the centre, then

In ∆POA, OA2 = OP2 + PA2

⇒ r2 = 42 + 32 = 52

⇒ r = 5

In ∆QOC, OC2 = CQ2 + OQ2

⇒ OQ2 = 42 + OQ2

⇒ OQ2 = 52 – 42 = 9

⇒ OQ = 3

**Question 4.**

Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Solution:

Given : ∠ABC is such that when we produce arms BA and BC, they make two equal chords AD and CE.

To prove: ∠ABC = [∠DOE – ∠AOC]

Construction: Join AE.

Proof: An exterior angle of a triangle is equal to the sum of interior opposite angles.

∴ In ∆BAE, we have

∠DAE = ∠ABC + ∠AEC ……(i)

The chord DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle.

⇒ ∠ABC = [(Angle subtended by the chord DE at the centre) – (Angle subtended by the chord AC at the centre)]

⇒ ∠ABC = [Difference of the angles subtended by the chords DE and AC at the centre]

Given : ∠ABC is such that when we produce arms BA and BC, they make two equal chords AD and CE.

To prove: ∠ABC = [∠DOE – ∠AOC]

Construction: Join AE.

Proof: An exterior angle of a triangle is equal to the sum of interior opposite angles.

∴ In ∆BAE, we have

∠DAE = ∠ABC + ∠AEC ……(i)

The chord DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle.

⇒ ∠ABC = [(Angle subtended by the chord DE at the centre) – (Angle subtended by the chord AC at the centre)]

⇒ ∠ABC = [Difference of the angles subtended by the chords DE and AC at the centre]

**Question 5.**

Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Solution:

We have a rhombus ABCD such that its diagonals AC and BD intersect at O.

Taking AB as diameter, a circle is drawn. Let us draw PQ || DA and RS || AB, both are passing through O.

P, Q, R and S are the mid-points of DC, AB, AD and BC respectively,

∵ Q is the mid-point of AB.

⇒ AQ = QB …(i)

Since AD = BC [ ∵ ABCD is a rhombus]

∴ AD = BC

⇒ RA = SB

⇒ RA = OQ …(ii)

[ ∵ PQ is drawn parallel to AD and AD = BC]

We have, AB = AD [Sides of rhombus are equal]

⇒ AB = AD

⇒ AQ = AR …(iii)

From (i), (ii) and (iii), we have AQ = QB = OQ

i.e. A circle drawn with Q as centre, will pass through A, B and O.

Thus, the circle passes through the of intersection ‘O’ of the diagonals rhombus ABCD.

We have a rhombus ABCD such that its diagonals AC and BD intersect at O.

Taking AB as diameter, a circle is drawn. Let us draw PQ || DA and RS || AB, both are passing through O.

P, Q, R and S are the mid-points of DC, AB, AD and BC respectively,

∵ Q is the mid-point of AB.

⇒ AQ = QB …(i)

Since AD = BC [ ∵ ABCD is a rhombus]

∴ AD = BC

⇒ RA = SB

⇒ RA = OQ …(ii)

[ ∵ PQ is drawn parallel to AD and AD = BC]

We have, AB = AD [Sides of rhombus are equal]

⇒ AB = AD

⇒ AQ = AR …(iii)

From (i), (ii) and (iii), we have AQ = QB = OQ

i.e. A circle drawn with Q as centre, will pass through A, B and O.

Thus, the circle passes through the of intersection ‘O’ of the diagonals rhombus ABCD.

**Question 6.**

ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.

Solution:

We have a circle passing through A, B and C is drawn such that it intersects CD at E.

ABCE is a cyclic quadrilateral.

∴∠AEC + ∠B = 180° …(i)

[Opposite angles of a cyclic quadrilateral are supplementary] But ABCD is a parallelogram. [Given]

∴∠D = ∠B …(ii)

[Opposite angles of a parallelogram are equal]

From (i) and (ii), we have

∠AEC + ∠D = 180° …(iii)

But ∠AEC + ∠AED = 180° [Linear pair] …(iv)

From (iii) and (iv), we have ∠D = ∠AED

i.e., The base angles of AADE are equal.

∴ Opposite sides must be equal.

⇒ AE = AD

We have a circle passing through A, B and C is drawn such that it intersects CD at E.

ABCE is a cyclic quadrilateral.

∴∠AEC + ∠B = 180° …(i)

[Opposite angles of a cyclic quadrilateral are supplementary] But ABCD is a parallelogram. [Given]

∴∠D = ∠B …(ii)

[Opposite angles of a parallelogram are equal]

From (i) and (ii), we have

∠AEC + ∠D = 180° …(iii)

But ∠AEC + ∠AED = 180° [Linear pair] …(iv)

From (iii) and (iv), we have ∠D = ∠AED

i.e., The base angles of AADE are equal.

∴ Opposite sides must be equal.

⇒ AE = AD

**Question 7.**

AC and BD are chords of a circle which bisect each other. Prove that

(i) AC and BD are diameters,

(ii) ABCD is a rectangle.AC and BD are chords of a circle which bisect each other. Prove that

(i) AC and BD are diameters,

(ii) ABCD is a rectangle.

Solution:

Given: A circle in which two chords AC and BD are such that they bisect each other. Let their point’of intersection be O.

To Prove: (i) AC and BD are diameters.

(ii) ABCD is a rectangle.

Construction: Join AB, BC, CD and DA.

Proof: (i) In ∆AOB and ∆COD, we have

AO = CO [O is the mid-point of AC]

BO = DO [O is the mid-point of BD]

∠AOB = ∠COD [Vertically opposite angles]

∴ Using the SAS criterion of congruence,

∆AOB ≅ ∆COD

⇒ AB = CD [C.P.C.T.]

⇒ arc AB = arc CD …(1)

Similarly, arc AD = arc BC …(2)

Adding (1) and (2), we get

arc AB + arc AD = arc CD + arc BC

⇒

⇒ BD divides the circle into two equal parts.

∴ BD is a diameter.

Similarly, AC is a diameter.

(ii) We know that ∆AOB ≅ ∆COD

⇒ ∠OAB = ∠OCD [C.P.C.T]

⇒ ∠CAB = ∠ACD

AB || DC

Similarly, AD || BC

∴ ABCD is a parallelogram.

Since, opposite angles of a parallelogram are equal.

∴ ∠DAB = ∠DCB

But ∠DAB + ∠DCB = 180°

[Sum of the opposite angles of a cyclic quadrilateral is 180°]

⇒ ∠DAB = 90° = ∠DCB Thus, ABCD is a rectangle.

Given: A circle in which two chords AC and BD are such that they bisect each other. Let their point’of intersection be O.

To Prove: (i) AC and BD are diameters.

(ii) ABCD is a rectangle.

Construction: Join AB, BC, CD and DA.

Proof: (i) In ∆AOB and ∆COD, we have

AO = CO [O is the mid-point of AC]

BO = DO [O is the mid-point of BD]

∠AOB = ∠COD [Vertically opposite angles]

∴ Using the SAS criterion of congruence,

∆AOB ≅ ∆COD

⇒ AB = CD [C.P.C.T.]

⇒ arc AB = arc CD …(1)

Similarly, arc AD = arc BC …(2)

Adding (1) and (2), we get

arc AB + arc AD = arc CD + arc BC

⇒

⇒ BD divides the circle into two equal parts.

∴ BD is a diameter.

Similarly, AC is a diameter.

(ii) We know that ∆AOB ≅ ∆COD

⇒ ∠OAB = ∠OCD [C.P.C.T]

⇒ ∠CAB = ∠ACD

AB || DC

Similarly, AD || BC

∴ ABCD is a parallelogram.

Since, opposite angles of a parallelogram are equal.

∴ ∠DAB = ∠DCB

But ∠DAB + ∠DCB = 180°

[Sum of the opposite angles of a cyclic quadrilateral is 180°]

⇒ ∠DAB = 90° = ∠DCB Thus, ABCD is a rectangle.

**Question 8.**

Bisectors of angles A, B and C of a ∆ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the ∆DEF are 90° – A, 90° – B and 90° – C.Bisectors of angles A, B and C of a ∆ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the ∆DEF are 90° – A, 90° – B and 90° – C.

Solution:

Given : A triangle ABC inscribed in a drcle, such that bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, £ and F respectively.

Construction: Join DE, EF and FD.

Proof:

∵ Angles in the same segment are equal.

∴ ∠ED A = ∠FCA …(i)

∠EDA = ∠EBA …(ii)

Adding (i) and (ii), we have

∠FDA + ∠EDA = ∠FCA + ∠EBA

⇒ ∠FDE = ∠FCA + ∠EBA

Given : A triangle ABC inscribed in a drcle, such that bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, £ and F respectively.

Construction: Join DE, EF and FD.

Proof:

∵ Angles in the same segment are equal.

∴ ∠ED A = ∠FCA …(i)

∠EDA = ∠EBA …(ii)

Adding (i) and (ii), we have

∠FDA + ∠EDA = ∠FCA + ∠EBA

⇒ ∠FDE = ∠FCA + ∠EBA

**Question 9.**

Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.

Solution:

We have two congruent circles such that they intersect each other at A and B. A line segment passing through A, meets the circles at P and Q.

Let us draw the common chord AB.

Since angles subtended by equal chords in the congruent circles are equal.

⇒ ∠APB = ∠AQB

Now, in ∆PBQ, we have ∠AQB = ∠APB

So, their opposite sides must be equal.

⇒ BP = BQ.

We have two congruent circles such that they intersect each other at A and B. A line segment passing through A, meets the circles at P and Q.

Let us draw the common chord AB.

Since angles subtended by equal chords in the congruent circles are equal.

⇒ ∠APB = ∠AQB

Now, in ∆PBQ, we have ∠AQB = ∠APB

So, their opposite sides must be equal.

⇒ BP = BQ.

**Question 10.**

In any ∆ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the ∆ABC.In any ∆ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the ∆ABC.

Solution:

∆ABC with O as centre of its circumcirde. The perpendicular bisector of BC passes through O. Join OB and OC. Suppose it cuts circumcirde at P.

In order to prove that the perpendicular bisector of BC and bisector of angle A of ∆ABC intersect at P, it is sufficient to show that AP is bisector of ∠A of ∆ABC.

Arc BC makes angle θ at the circumference

∴ ∠BOC = 2θ

[Angle at centre is double the angle made by an arc at circumference]

Also, in ∆BOC, OB=OC and OP is perpendicular bisector of BC.

So, ∠BOP = ∠COP = θ

Arc CP makes angle θ at O, so it will make

angle [/latex]\frac { \theta }{ 2 }[/latex] at circumference.

So, ∠COP = [/latex]\frac { \theta }{ 2 }[/latex]

Hence, AP is the angle bisedor of ∠A of ∆ABC.

In order to prove that the perpendicular bisector of BC and bisector of angle A of ∆ABC intersect at P, it is sufficient to show that AP is bisector of ∠A of ∆ABC.

Arc BC makes angle θ at the circumference

∴ ∠BOC = 2θ

[Angle at centre is double the angle made by an arc at circumference]

Also, in ∆BOC, OB=OC and OP is perpendicular bisector of BC.

So, ∠BOP = ∠COP = θ

Arc CP makes angle θ at O, so it will make

angle [/latex]\frac { \theta }{ 2 }[/latex] at circumference.

So, ∠COP = [/latex]\frac { \theta }{ 2 }[/latex]

Hence, AP is the angle bisedor of ∠A of ∆ABC.