# NCERT Solutions for Class 9 Maths Chapter 10 - Circles

## Chapter 10 - Circles Exercise Ex. 10.1

Solution 1

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

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

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

the circle. (exterior/interior)

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

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

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

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

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

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

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

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

Solution 2

(i) True, all the points on circle are at equal distance from the centre of circle, and this equal distance

it called as radius of circle.

(ii) False, on a circle there are infinite points. So, we can draw infinite number of chords of given length. Hence, a circle has infinite number of equal chords.

(iii) False, consider three arcs of same length as AB, BC and CA. Now we may observe that for minor arc BDC. CAB is major arc. So AB, BC and CA are minor arcs of circle.

(iv) True, let AB be a chord which is twice as long as its radius. In this situation our chord will be passing through centre of circle. So it will be the diameter of circle.

(v) False, sector is the region between an arc and two radii joining the centre to the end points of the arc as in the given figure OAB is the sector of circle.

(vi) True, A circle is a two dimensional figure and it can also be referred as plane figure.

## Chapter 10 - Circles Exercise Ex. 10.2

Solution 1

A circle is a collection of points which are equidistant from a fix point. This fix point is called as the centre of circle and this equal distance is called as radius of circle. And thus shape of a circle depends on the radius of the circle.

So, if we try to superimpose two circles of equal radius, one each other both circles will cover each other.

So, two circles are congruent if they have equal radius.

So, if we try to superimpose two circles of equal radius, one each other both circles will cover each other.

So, two circles are congruent if they have equal radius.

Now consider two congruent circles having centre O and O' and two chords AB and CD of equal lengths

Now in AOB and CO'D

AB = CD (chords of same length)

OA = O'C (radii of congruent circles)

OB = O'D (radii of congruent circles)

AOB CO'D (SSS congruence rule)

AB = CD (chords of same length)

OA = O'C (radii of congruent circles)

OB = O'D (radii of congruent circles)

AOB CO'D (SSS congruence rule)

AOB = CO'D (by CPCT)

Hence equal chords of congruent circles subtend equal angles at their centres.

Solution 2

Let us consider two congruent circles (circles of same radius) with centres as O and O'.

In AOB and CO'D

AOB = CO'D (given)

OA = O'C (radii of congruent circles)

OB = O'D (radii of congruent circles)

AOB CO'D (SSS congruence rule)

AB = CD (by CPCT)

Hence, if chords of congruent circles subtend equal angles at their centres then chords are equal.

AOB = CO'D (given)

OA = O'C (radii of congruent circles)

OB = O'D (radii of congruent circles)

AOB CO'D (SSS congruence rule)

AB = CD (by CPCT)

Hence, if chords of congruent circles subtend equal angles at their centres then chords are equal.

## Chapter 10 - Circles Exercise Ex. 10.3

Solution 1

Consider the following pair of circles.

(i) circles don't intersect each other at any point, so circles are not having any point in common.

(i) circles don't intersect each other at any point, so circles are not having any point in common.

(ii) Circles touch each other only at one point P so there is only 1 point in common.

(iii) Circles touch each other at 1 point X only. So the circles have 1 point in common.

(iv) These circles intersect each other at two points P and Q. So the circles have two points in common. We may observe that there can be maximum 2 points in common.

(iv) These circles intersect each other at two points P and Q. So the circles have two points in common. We may observe that there can be maximum 2 points in common.

We can have a situation in which two congruent circles are superimposed on each other, this situation can be referred as if we are drawing circle two times.

Solution 2

Following are the steps of construction:

Step1. Take the given circle centered at point O.

Step2. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these

Step1. Take the given circle centered at point O.

Step2. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these

chords.

Step3. Let these perpendicular bisectors meet at point O. Now, O is the centre of given circle.

Step3. Let these perpendicular bisectors meet at point O. Now, O is the centre of given circle.

Solution 3

Consider two circles centered at point O and O' intersect each other at point A and B respectively.

Join AB. AB is the chord for circle centered at O, so perpendicular bisector of AB will pass through O.

Again AB is also chord of circle centered at O', so, perpendicular bisector of AB will also pass through O'.

Clearly centres of these circles lie on the perpendicular bisector of common chord.

Join AB. AB is the chord for circle centered at O, so perpendicular bisector of AB will pass through O.

Again AB is also chord of circle centered at O', so, perpendicular bisector of AB will also pass through O'.

Clearly centres of these circles lie on the perpendicular bisector of common chord.

## Chapter 10 - Circles Exercise Ex. 10.4

Solution 1

Let radius of circle centered at O and O' be 5 cm and 3 cm respectively.

OA = OB = 5 cm

O'A = O'B = 3 cm

OO' will be the perpendicular bisector of chord AB.

AC = CB

OA = OB = 5 cm

O'A = O'B = 3 cm

OO' will be the perpendicular bisector of chord AB.

AC = CB

Given that OO' = 4 cm

Let OC be x. so, O'C will be 4 - x

In OAC

OA

^{2}= AC

^{2}+ OC

^{2}

5

^{2}= AC

^{2}+ x

^{2}

25 - x

^{2}= AC

^{2}... (1)

In O'AC

O'A

^{2}= AC

^{2}+ O'C

^{2}

3

^{2}= AC

^{2}+ (4 - x)

^{2}

9 = AC

^{2}+ 16 + x

^{2}- 8x

AC

^{2}= - x

^{2}- 7 + 8x ... (2)

From equations (1) and (2), we have

25 - x

^{2}= - x

^{2}- 7 + 8x

8x = 32

x = 4

So, the common chord will pass through the centre of smaller circle i.e. O'. and hence it will be diameter of smaller circle.

Now, AC

AC = 3 m

^{2}= 25 - x^{2}= 25 - 4^{2}= 25 - 16 = 9AC = 3 m

The length of the common chord AB = 2 AC = (2 3) m = 6 m

Solution 2

Let PQ and RS are two equal chords of a given circle and there are intersecting each other at point T.

Draw perpendiculars OV and OU on these chords.

In OVT and OUT

OV = OU (Equal chords of a circle are equidistant from the centre)

OVT = OUT (Each 90

OT = OT (common)

In OVT and OUT

OV = OU (Equal chords of a circle are equidistant from the centre)

OVT = OUT (Each 90

^{o})OT = OT (common)

OVT OUT (RHS congruence rule)

VT = UT (by CPCT) ... (1)

It is given that

PQ = RS ... ... ... ... (2)

PV = RU ... ... ... ... (3)

On adding equations (1) and (3), we have

PV + VT = RU + UT

PT = RT ... ... ... ... (4)

On subtracting equation (4) from equation (2), we have

PQ - PT = RS - RT

QT = ST ... ... ... ... (5)

Equations (4) and (5) shows that the corresponding segments of

chords PQ and RS are congruent to each other.

On adding equations (1) and (3), we have

PV + VT = RU + UT

PT = RT ... ... ... ... (4)

On subtracting equation (4) from equation (2), we have

PQ - PT = RS - RT

QT = ST ... ... ... ... (5)

Equations (4) and (5) shows that the corresponding segments of

chords PQ and RS are congruent to each other.

Solution 3

Let PQ and RS are two equal chords of a given circle and there are intersecting each other at point T.

Draw perpendiculars OV and OU on these chords.

In OVT and OUT

OV = OU (Equal chords of a circle are equidistant from the centre)

OVT = OUT (Each 90

OT = OT (common)

OVT OUT (RHS congruence rule)

OTV = OTU (by CPCT)

Draw perpendiculars OV and OU on these chords.

In OVT and OUT

OV = OU (Equal chords of a circle are equidistant from the centre)

OVT = OUT (Each 90

^{o})OT = OT (common)

OVT OUT (RHS congruence rule)

OTV = OTU (by CPCT)

Hence, the line joining the point of intersection to the centre makes equal angles with the chords.

Solution 4

Let us draw a perpendicular OM on line AD.

Here, BC is chord of smaller circle and AD is chord of bigger circle.

We know that the perpendicular drawn from centre of circle bisects the chord.

BM = MC ... (1)

And AM = MD ... (2)

Subtracting equations (2) from (1), we have

AM - BM = MD - MC

AB = CD

Solution 5

Draw perpendiculars OA and OB on RS and SM respectively.

Let R, S and M be the position of Reshma, Salma and Mandip respectively.

Let R, S and M be the position of Reshma, Salma and Mandip respectively.

AR = AS = = 3cm

OR = OS = OM = 5 m (radii of circle)

In OAR

OA

OA

OA

OA = 4 m

We know that in an isosceles triangle altitude divides the base, so in RSM

RCS will be of 90

Area of ORS = OARS

In OAR

OA

^{2}+ AR^{2}= OR^{2}OA

^{2}+ (3 m)^{2}= (5 m)^{2}OA

^{2}= (25 - 9) m^{2}= 16 m^{2}OA = 4 m

We know that in an isosceles triangle altitude divides the base, so in RSM

RCS will be of 90

^{o}and RC = CMArea of ORS = OARS

RC = 4.8

RM = 2RC = 2(4.8)= 9.6

So, distance between Reshma and Mandip is 9.6 m.

Solution 6

Given that AS = SD = DA

So, ASD is a equilateral triangle

OA (radius) = 20 m.

Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC.

We also know that median intersect each other at the 2: 1. As AB is the median of equilateral triangle ABC, we can write

So, ASD is a equilateral triangle

OA (radius) = 20 m.

Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC.

We also know that median intersect each other at the 2: 1. As AB is the median of equilateral triangle ABC, we can write

AB = OA + OB = (20 + 10) m = 30 m.

In ABD

AD

^{2}= AB

^{2}+ BD

^{2}

AD

^{2}= (30)

^{2}+

So, length of string of each phone will be m.

## Chapter 10 - Circles Exercise Ex. 10.5

Solution 1

We may observe that

AOC = AOB + BOC

= 60

= 90

We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.

AOC = AOB + BOC

= 60

^{o}+ 30^{o}= 90

^{o}We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.

Solution 2

In OAB

AB = OA = OB = radius

OAB is an equilateral triangle.

AB = OA = OB = radius

OAB is an equilateral triangle.

So, each interior angle of this triangle will be of 60

^{o}

AOB = 60

^{o}Now,

In cyclic quadrilateral ACBD

ACB + ADB = 180

ADB = 180

So, angle subtended by this chord at a point on major arc and minor arc are 30

ACB + ADB = 180

^{o}(Opposite angle in cyclic quadrilateral)ADB = 180

^{o}- 30^{o}= 150^{o}So, angle subtended by this chord at a point on major arc and minor arc are 30

^{o}and 150^{o}respectively.Solution 3

Consider PR as a chord of circle.

Take any point S on major arc of circle.

Now PQRS is a cyclic quadrilateral.

PQR + PSR = 180

PSR = 180

We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.

POR = 2PSR = 2 (80

^{o}(Opposite angles of cyclic quadrilateral)PSR = 180

^{o}- 100^{o}= 80^{o}We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.

POR = 2PSR = 2 (80

^{o}) = 160^{o}In POR

OP = OR (radii of same circle)

OP = OR (radii of same circle)

OPR = ORP (Angles opposite equal sides of a triangle)

OPR + ORP + POR = 180

^{o}(Angle sum property of a triangle)2 OPR + 160

2 OPR = 180

^{o}= 180^{o}2 OPR = 180

^{o}- 160^{o}= 20^{o}OPR = 10

^{o}Solution 4

In ABC

BAC + ABC + ACB = 180

BAC + 69

^{o}(Angle sum property of a triangle)BAC + 69

^{o}+ 31^{o}= 180^{o}BAC = 180

BAC = 80

^{o}- 100ºBAC = 80

^{o}BDC = BAC = 80

^{o}(Angles in same segment of circle are equal)Solution 5

In CDE

CDE + DCE = CEB (Exterior angle)

CDE + 20

CDE + DCE = CEB (Exterior angle)

CDE + 20

^{o}= 130^{o}CDE = 110

But BAC = CDE (Angles in same segment of circle)

BAC = 110

^{o}But BAC = CDE (Angles in same segment of circle)

BAC = 110

^{o}Solution 6

For chord CD

CBD = CAD (Angles in same segment)

CAD = 70

^{o}BAD = BAC + CAD = 30

BCD + BAD = 180

BCD + 100

BCD = 80

In ABC

^{o}+ 70^{o}= 100^{o}BCD + BAD = 180

^{o}(Opposite angles of a cyclic quadrilateral)BCD + 100

^{o}= 180^{o}BCD = 80

^{o}In ABC

AB = BC (given)

BCA = CAB (Angles opposite to equal sides of a triangle)

BCA = 30

We have BCD = 80

BCA + ACD = 80

30

BCA = CAB (Angles opposite to equal sides of a triangle)

BCA = 30

^{o}We have BCD = 80

^{o}BCA + ACD = 80

^{o}30

^{o}+ ACD = 80^{o}ACD = 50

^{o}ECD = 50

^{o}Solution 7

Let ABCD a cyclic quadrilateral having diagonals as BD and AC intersecting each other at point O.

(Consider BD as a chord)

BCD + BAD = 180

BCD = 180

^{o}(Cyclic quadrilateral)BCD = 180

^{o}- 90^{o}= 90^{o} (Considering AC as a chord)

ADC + ABC = 180

90

ABC = 90

Here, each interior angle of cyclic quadrilateral is of 90

^{o}(Cyclic quadrilateral)90

^{o}+ ABC = 180^{o}ABC = 90

^{o}Here, each interior angle of cyclic quadrilateral is of 90

^{o}. Hence it is a rectangle.Solution 8

Consider a trapezium ABCD with AB | |CD and BC = AD Draw AM CD and BN CD

In AMD and BNC

AD = BC (Given)

AMD = BNC (By construction each is 90

AM = BM (Perpendicular distance between two parallel lines is same)

AMD BNC (RHS congruence rule)

In AMD and BNC

AD = BC (Given)

AMD = BNC (By construction each is 90

^{o})AM = BM (Perpendicular distance between two parallel lines is same)

AMD BNC (RHS congruence rule)

ADC = BCD (CPCT) ... (1)

BAD and ADC are on same side of transversal AD

BAD + ADC = 180

BAD + BCD = 180

This equation shows that the opposite angles are supplementary.

So, ABCD is a cyclic quadrilateral.

^{o}... (2)BAD + BCD = 180

^{o}[Using equation (1)]This equation shows that the opposite angles are supplementary.

So, ABCD is a cyclic quadrilateral.

Solution 9

Join chords AP and DQ

For chord AP

PBA = ACP (Angles in same segment) ... (1)

For chord DQ

DBQ = QCD (Angles in same segment) ... (2)

ABD and PBQ are line segments intersecting at B.

PBA = DBQ (Vertically opposite angles) ... (3)

From equations (1), (2) and (3), we have

ACP = QCD

For chord AP

PBA = ACP (Angles in same segment) ... (1)

For chord DQ

DBQ = QCD (Angles in same segment) ... (2)

ABD and PBQ are line segments intersecting at B.

PBA = DBQ (Vertically opposite angles) ... (3)

From equations (1), (2) and (3), we have

ACP = QCD

Solution 10

Consider a ABC

Two circles are drawn while taking AB and AC as diameter.

Let they intersect each other at D and let D does not lie on BC.

Join AD

ADB = 90

ADC = 90

BDC = ADB + ADC = 90

Hence BDC is straight line and our assumption was wrong.

Thus, Point D lies on third side BC of ABC

Two circles are drawn while taking AB and AC as diameter.

Let they intersect each other at D and let D does not lie on BC.

Join AD

ADB = 90

^{o}(Angle subtend by semicircle)ADC = 90

^{o}(Angle subtend by semicircle)BDC = ADB + ADC = 90

^{o}+ 90^{o}= 180^{o}Hence BDC is straight line and our assumption was wrong.

Thus, Point D lies on third side BC of ABC

Solution 11

In ABC

ABC + BCA + CAB = 180

90

BCA + CAB = 90

In ADC

^{o}(Angle sum property of a triangle)90

^{o}+ BCA + CAB = 180^{o}BCA + CAB = 90

^{o}... (1)In ADC

CDA + ACD + DAC = 180

90

ACD + DAC = 90

Adding equations (1) and (2), we have

^{o}(Angle sum property of a triangle)90

^{o}+ ACD + DAC = 180^{o}ACD + DAC = 90

^{o}... (2)Adding equations (1) and (2), we have

BCA + CAB + ACD + DAC = 180

^{o} (BCA + ACD) + (CAB + DAC) = 180

But it is given that

B + D = 90

From equations (3) and (4), we can see that quadrilateral ABCD is having sum of measures of opposite angles as 180

So, it is a cyclic quadrilateral.

Consider chord CD.

Now, CAD = CBD (Angles in same segment)

^{o}BCD + DAB = 180^{o}... (3)But it is given that

B + D = 90

^{o}+ 90^{o}= 180^{o}... (4)From equations (3) and (4), we can see that quadrilateral ABCD is having sum of measures of opposite angles as 180

^{o}.So, it is a cyclic quadrilateral.

Consider chord CD.

Now, CAD = CBD (Angles in same segment)

Solution 12

Let ABCD be a cyclic parallelogram.

A + C = 180

We know that opposite angles of a parallelogram are equal

A = C and B = D

From equation (1)

A + C = 180

A + A = 180

2 A = 180

A = 90

Parallelogram ABCD is having its one of interior angles as 90

A + C = 180

^{o}(Opposite angle of cyclic quadrilateral) ... (1)We know that opposite angles of a parallelogram are equal

A = C and B = D

From equation (1)

A + C = 180

^{o}A + A = 180

^{o}2 A = 180

^{o}A = 90

^{o}Parallelogram ABCD is having its one of interior angles as 90

^{o}, so, it is a rectangle.## Chapter 10 - Circles Exercise Ex. 10.6

Solution 1

Let two circles having their centres as O and intersect each other at point A and B respectively.

Construction: Let us join OO',

Construction: Let us join OO',

Now in AOO' and BOO'

OA = OB (radius of circle 1)

O'A = O'B (radius of circle 2)

OO' = OO' (common)

OA = OB (radius of circle 1)

O'A = O'B (radius of circle 2)

OO' = OO' (common)

AOO' BOO' (by SSS congruence rule)

OAO' = OBO' (by CPCT)

So, line of centres of two intersecting circles subtends equal angles at the two points of intersection.

OAO' = OBO' (by CPCT)

So, line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Solution 2

Draw OM AB and ON CD. Join OB and OD

(Perpendicular from centre bisects the chord)

Let ON be x, so OM will be 6 - x

In MOB

In MOB

In NOD

We have OB = OD (radii of same circle)

So, from equation (1) and (2)

So, from equation (1) and (2)

From equation (2)

So, radius of circle is found to be cm.

Solution 3

Distance of smaller chord AB from centre of circle = 4 cm.

OM = 4 cm

OM = 4 cm

In OMB

In OND

OD=OB=5cm (radii of same circle)

So, distance of bigger chord from centre is 3 cm.

Solution 4

In AOD and COE

OA = OC (radii of same circle)

OD = OE (radii of same circle)

AD = CE (given)

AOD COE (SSS congruence rule)

OA = OC (radii of same circle)

OD = OE (radii of same circle)

AD = CE (given)

AOD COE (SSS congruence rule)

OAD = OCE (by CPCT) ... (1)

ODA = OEC (by CPCT) ... (2)

We also have

OAD = ODA (As OA = OD) ... (3)

From equations (1), (2) and (3), we have

OAD = OCE = ODA = OEC

Let OAD = OCE = ODA = OEC = x

In OAC,

OA = OC

OCA = OAC (let a)

ODA = OEC (by CPCT) ... (2)

We also have

OAD = ODA (As OA = OD) ... (3)

From equations (1), (2) and (3), we have

OAD = OCE = ODA = OEC

Let OAD = OCE = ODA = OEC = x

In OAC,

OA = OC

OCA = OAC (let a)

In ODE,

OD = OE

OED = ODE (let y)

ADEC is a cyclic quadrilateral

CAD + DEC = 180

OD = OE

OED = ODE (let y)

ADEC is a cyclic quadrilateral

CAD + DEC = 180

^{o}(opposite angles are supplementary)x + a + x + y = 180

2x + a + y = 180

y = 180 - 2x - a ... (4)

But DOE = 180 - 2y

And AOC = 180 - 2a

Now, DOE - AOC = 2a - 2y = 2a - 2 (180 - 2x - a)

= 4a + 4x - 360

Now, BAC + CAD = 180 (Linear pair)

BAC = 180 - CAD = 180 - (a + x)

Similarly, ACB = 180 - (a + x)

Now, in ABC

ABC + BAC + ACB = 180 (Angle sum property of a triangle)

ABC = 180 - BAC - ACB

= 180 - (180 - a - x) - (180 - a -x)

= 2a + 2x - 180

= [4a + 4x - 360

ABC = [DOE - AOC] [Using equation (5)]

^{o}2x + a + y = 180

^{o}y = 180 - 2x - a ... (4)

But DOE = 180 - 2y

And AOC = 180 - 2a

Now, DOE - AOC = 2a - 2y = 2a - 2 (180 - 2x - a)

= 4a + 4x - 360

^{o}... (5)Now, BAC + CAD = 180 (Linear pair)

BAC = 180 - CAD = 180 - (a + x)

Similarly, ACB = 180 - (a + x)

Now, in ABC

ABC + BAC + ACB = 180 (Angle sum property of a triangle)

ABC = 180 - BAC - ACB

= 180 - (180 - a - x) - (180 - a -x)

= 2a + 2x - 180

= [4a + 4x - 360

^{o}]ABC = [DOE - AOC] [Using equation (5)]

Solution 5

Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn taking side CD as its diameter.

We know that angle in a semicircle is of 90

COD = 90

We know that angle in a semicircle is of 90

^{o}.COD = 90

^{o}Also in rhombus the diagonals intersect each other at 90

^{o}

AOB = BOC = COD = DOA = 90

^{o}

So, point O has to lie on the circle.

Solution 6

We see that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral sum of opposite angles is 180

AEC + CBA = 180

AEC + AED = 180

AED = CBA ... (1)

For a parallelogram opposite angles are equal.

ADE = CBA ... (2)

From (1) and (2)

AED = ADE

AD = AE (angles opposite to equal sides of a triangle)

^{o}AEC + CBA = 180

^{o}AEC + AED = 180

^{o}(linear pair)AED = CBA ... (1)

For a parallelogram opposite angles are equal.

ADE = CBA ... (2)

From (1) and (2)

AED = ADE

AD = AE (angles opposite to equal sides of a triangle)

Solution 7

Let two chords AB and CD are intersecting each other at point O.

In AOB and COD

OA = OC (given)

OB = OD (given)

AOB = COD (vertically opposite angles)

AOB COD (SAS congruence rule)

AB = CD (by CPCT)

Similarly, we can prove AOD COB

AD = CB (by CPCT)

In AOB and COD

OA = OC (given)

OB = OD (given)

AOB = COD (vertically opposite angles)

AOB COD (SAS congruence rule)

AB = CD (by CPCT)

Similarly, we can prove AOD COB

AD = CB (by CPCT)

Since in quadrilateral ACBD opposite sides are equal in length.

Hence, ACBD is a parallelogram.

We know that opposite angles of a parallelogram are equal

A = C

Hence, ACBD is a parallelogram.

We know that opposite angles of a parallelogram are equal

A = C

But A + C = 180

A + A = 180

A = 180

A = 90

As ACBD is a parallelogram and one of its interior angles is 90

A is the angle subtended by chord BD. And as A = 90

^{o}(ABCD is a cyclic quadrilateral)A + A = 180

^{o}A = 180

^{o}A = 90

^{o}As ACBD is a parallelogram and one of its interior angles is 90

^{o}, so it is a rectangle.A is the angle subtended by chord BD. And as A = 90

^{o}, so BD should be diameter of circle. Similarly AC is diameter of circle.Solution 8

It is given that BE is the bisector of B

ABE =

But ADE = ABE (angles in same segment for chord AE)

ADE =

Similarly, ACF = ADF = (angle in same segment for chord AF)

Now, D = ADE + ADF

Similarly we can prove that

Solution 9

AB is common chord in both congruent circles.

APB = AQB

APB = AQB

Now in BPQ

APB = AQB

BP = BQ (angles opposite to equal sides of a triangle)

Solution 10

Let perpendicular bisector of side BC and angle bisector of A meet at point D.

Let perpendicular bisector of side BC intersects it at E.

Perpendicular bisector of side BC will pass through circum centre O of circle. Now, BOC and BAC are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively.

We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

BOC = 2 BAC = 2A ... (1)

In BOE and COE

Let perpendicular bisector of side BC intersects it at E.

Perpendicular bisector of side BC will pass through circum centre O of circle. Now, BOC and BAC are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively.

We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

BOC = 2 BAC = 2A ... (1)

In BOE and COE

OE = OE (common)

OB = OC (radii of same circle)

OB = OC (radii of same circle)

OEB = OEC (Each 90

^{o}as OD BC) BOE COE (RHS congruence rule)

BOE = COE (by CPCT) ... (2)

But BOE + COE = BOC

But BOE + COE = BOC

BOE +BOE = 2 A [Using equations (1) and (2)]

BOE = 2A

BOE = 2A

BOE = A

BOE = COE = A

The perpendicular bisector of side BC and angle bisector of A meet at point D.

BOD = BOE = A ... (3)

Since AD is the bisector of angle A

BAD =

2BAD = A ... (4)

From equations (3) and (4), we have

BOD = 2 BAD

It is possible only if BD will be a chord of the circle. For this the point D lies on circum circle.

Therefore, the perpendicular bisector of side BC and angle bisector of A meet on the circum circle of triangle ABC.

2BAD = A ... (4)

From equations (3) and (4), we have

BOD = 2 BAD

It is possible only if BD will be a chord of the circle. For this the point D lies on circum circle.

Therefore, the perpendicular bisector of side BC and angle bisector of A meet on the circum circle of triangle ABC.

### Kindly Sign up for a personalised experience

- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions

#### Sign Up

#### Verify mobile number

Enter the OTP sent to your number

Change