Chapter 11 Circles

**NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.1**

**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.**

**Solution:**

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

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

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

**(iv)** An arc is a semi-circle 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.

**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.**

**Solution:**

**(i) True.** Because all points are equidistant from the centre to the circle.

**(ii) False.** Because circle has infinitely may equal chords can be drawn.

**(iii) False.** Because all three arcs are equal, so their is no difference between the major and minor arcs.

**(iv) True.** By the definition of diameter, that diameter is twice the radius.

**(v) False.** Because the sector is the region between two radii and an arc.

**(vi) True**. Because circle is a part of the plane figure.

**NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.2**

**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**

**Solution:**

Given MN and PQ are two equal chords of two congruent circles with centre at O and O’.

To prove ∠ MON = ∠ PO’Q

Proof In ∆ MON and ∆ PO’Q, we have

MO = PO’ (Radii of congruent circles)

NO = QO’ (Radii of congruent circles)

and MN = PQ (Given)

∴ By SSS criterion, we get

∆ MON = ∆ PO’Q

Hence, ∠ MON = ∠ PO’Q (By CPCT)

**Question 2.**

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

**Solution:**

Given MN and PQ are two chords of congruent circles such that angles subtended by .

these chords at the centres O and O’ of the circles are equal.

To prove MN = PQ

Proof In ∆ MON and ∆ PO’Q, we get

MO = PO’ (Radii of congruent circles)

NO = QO’ (Radii of congruent circles)

and ∠MON = ∠PO’Q (Given)

∴ By SAS criteria, we get

∆ MON = ∆ PO’Q

Hence, MN = PQ (By CPCT)

**NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.3**

**Question 1.**

**Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?**

**Solution:**

Different pairs of circles are

**(i) Two points common**

**(ii) One point is common**

**(iii) No point is common**

**(iv) No point is common**

**(v) One point is common**

From figures, it is obvious that these pairs many have 0 or 1 or 2 points in common.

Hence, a pair of circles cannot intersect each other at more than two points.

**Question 2.**

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

**Solution:**

**Steps of construction**

Taking three points P,Q and R on the circle.

Join PQ and QR,

Draw MQ and NS, respectively the perpendicular bisectors of PQ and RQ, which intersect each other at O.

Hence, O is the centre of the circle.

**Question 3.**

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

**Solution:**

**Given:** Two circles with centres O and O’ intersect at two points M and N so that MN is the common chord of the two circles and OO’ is the line segment joining the centres of the two circles. Let OO’ intersect MN at P.

**To prove:** OO’ is the perpendicular bisector of MN.

**Construction:** Draw line segments OM, ON, O’M and O’N.

Proof In ∆ OMO’ and ONO’, we get

OM = ON (Radii of the same circle)

O’M = O’N (Radii of the same circle)

OO’ = OO’ (Common)

∴ By SSS criterion, we get

∆ OMO’ ≅ ONO’

So, ∠ MOO’ = ∠ N00′ (By CPCT)

∴ ∠ MOP = ∠ NOP …(i)

(∵ ∠ MOO’ = ∠ MOP and ∠ NOO’ = ∠ NOP)

In ∆ MOP and ∆ NOP, we get

OM = ON (Radii of the same circle)

∠ MOP = ∠NOP [ From Eq(i)]

and OM = OM (Common)

∴ By SAS criterion, we get

∆ MOP ≅ ∆NOP

So, MP = NP (By CPCT)

and ∠ MPO = ∠ NPO

But ∠ MPO + ∠NPO = 180° ( ∵MPN is a straight line)

∴ 2 ∠ MPO = 180° ( ∵ ∠ MPO = ∠ NPO)

⇒ ∠ MPO = 90°

So, MP = PN

and ∠ MPO = ∠ NPO = 90°

Hence, OO’ is the perpendicular bisector of MN.

**NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.4**

**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.**

**Solution:**

Let O and O’ be the centres of the circles of radii 5 cm and 3 cm, respectively.

Let AB be their common chord.

Given, OA = 5 cm, O’A = 3cm and OO’ = 4 cm

∴ AO’^{2} + OO’^{2} = 3^{2} + 4^{2} = 9 + 16- 25 = OA^{2}

∴ OO’A is a right angled triangle and right angled at O’

Area of ∆OO’A = 12 x O’A x OO’

= 12 x 3x 4= 6sq units …(i)

Also, area of ∆OO’A = 12 x OO’ x AM

= 12 x 4 x AM =2 AM …(ii)

From Eqs. (i) and (ii), we get

2AM = 6 ⇒ AM = 3

Since, when two circles intersect at two points, then their centre lie on the perpendicular bisector of the common chord.

∴ AB = 2 x AM= 2 x 3 = 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.**

**Solution:**

**Given:** MN and AS are two chords of a circle with centre O, AS and MN intersect at P and MN = AB

**To prove:** MP = PB and PN = AP

**Construction:** Draw OD ⊥ MN and OC ⊥ AB.

Join OP

**Proof :** ∵ DM = DN = 12 MN (Perpendicular from centre bisects the chord)

and AC = CB = 12 AB (Perpendicular from centre bisects the chord)

MD = BC and DN = AC (∵ MN = AS)…(i)

in ∆ODP and ∆OPC

OD = OC (Equal chords of a circle are equidistant from the centre)

∠ ODP = ∠OCP

OP = OP (Common)

∴ RHS criterion of congruence,

∆ ODP ≅ ∆ OCP

∴ DP = PC (By CPCT)…(ii)

On adding Eqs. (i) and (ii), we get

MD + DP = BC + PC

MP = PB

On subtracting Eq. (ii) from Eq. (i), we get

DN – DP = AC – PC

PN = AP

Hence, MP = PB and PN = AP are proved.

**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.**

**Solution:**

**Given:** RQ and MN are chords of a with centre O. MN and RQ intersect at P and MN = RQ

**To prove:** ∠ OPC = ∠ OPB

**Construction:** Draw OC ⊥ RQ and OB ⊥ MN.

Join OP.

**Proof:** In ∆ OCP and ∆ OBP, we get

∠ OCP = ∠ OBP (Each = 90°)

OP = OP (Common)

OC = OB (Equal chords of a circle are equidistant from the centre)

∴ By RHS criterion of congruence, we get

∆ OCP ≅ ∆ OBP

∴ ∠ OPC = ∠ OPB (By CPCT)

**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).**

**Solution:**

Let OP be the perpendicular from O on line l. Since, the perpendicular from the centre of a circle to a chord

Now, BC is the chord of the smaller circle and OP ⊥ BC.

∴ BP = PC ……(i)

Since, AD is a chord of the larger circle and OP ⊥ AD.

∴ AP = PD …(ii)

On subtracting Eq. (i) from Eq. (ii), we get

AP – BP = PD – PC

⇒ AB = CD

Hence proved.

**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?**

**Solution:**

Let O be the centre of the circle and Reshma, Salma and Mandip are represented by the points Ft, S and M, respectively.

Let RP = xm.

From Eqs. (i) and (ii), we get

Hence, the 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.**

**Solution:**

Let Ankur, Syed and David standing on the point P, Q and R.

Let PQ = QR = PR = x

Therefore, ∆ PQR is an equilateral triangle. Drawn altitudes PC, QD and RN from vertices to the sides of a triangle and intersect these altitudes at the centre of a circle M.

As PQR is an equilateral, therefore these altitudes bisects their sides.

In ∆ PQC,

PQ^{2} = PC^{2} + QC^{2} (By Pythagoras theorem)

**NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.5**

**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.**

**Solution:**

∴ ∠AOC = ∠AOB + ∠BOC = 60P + 30° = 90°

∴ Arc ABC makes 90° at the centre of the circle.

∴ ∠ADC = 12 ∠AOC

(∵ The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle.)

= 12 x 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.**

**Solution:**

Let BC be chord, which is equal to the radius. Join OB and OC.

Given, BC=OB = OC

∴ ∆OBC is an equilateral triangle.

∠BOC =60°

∴ BAC = 12 ∠BOC

= 12 x 60° = 30°

(∵ The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle.)

Here, ABMC is a cyclic quadrilateral.

∴ ∠BAC + ∠BMC = 180°

(∵ In a cyclic quadrilateral the sum of opposite angles is 180°)

⇒ ∠BMC= 180° – 30° =150°

**Question 3.**

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

**Solution:**

∴ ∠POR = 2∠PQR = 2 x 100° = 200°

(Since, the angle subtended by the centre is double the angle subtended by circumference.)

Since, in ∆OPR, ∠POR = 360° – 200° = 160° .. (i)

Again, ∆ OPR, OP = OR (Radii of the circle)

∴ ∠OPR = ∠ORP (By property of isosceles triangle)

In ∆POR, ∠OPR + ∠ORP + ∠POR = 180° …(ii)

From Eqs. (i) and (ii), we get

∠OPR + ∠OPR + 160° = 180°

∴ 2 ∠OPR = 180° – 160° = 20°

∴ ∠OPR = 120circ2 = 10°

**Question 4.**

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

**Solution:**

∵ ∠BDC = ∠BAC …(i)

(Since, the angles in the same segment are equal)

Now , in ∆ABC

∴ ∠A + ∠B+ ∠C= 180°

⇒ ∠A+ 69°+ 31° = 180°

⇒ ∠A + 100° = 180°

∴ ∠A = 180° – 100° = 80°

⇒ ∠BAC=80°

∴ From Eq.(i)∠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.**

**Solution:**

∴ ∠AEB = 180° – 130° = 50° (Linear Pair) …(i)

⇒ ∠CED = ∠AEB = 50° (Vertically opposite)

Again ∠ABD = ∠ACD (Since, the angles in the same segment are equal)

∠ABE = ∠ECD

⇒ ∠ABE = 180° …(ii)

∴ In ∆ CDE

∠A+ 20° + 50° = 180° [From Eqs. (i) and (ii)]

∠A + 70° = 180°

∴ ∠A = 180°- 70° =110°

Hence ∠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.**

**Solution:**

Angles in the same segment are equal.

∴ ∠BDC = ∠BAC

∴ ∠BDC = 30°

In ∆ BCD, we have

∴ ∠BDC + ∠DBC + ∠BCD = 180° (Given, ∠DBC = 70° and ∠BDC = 30°)

∴ 30° + 70° + ∠BCD = 180°

∴ ∠BCD= 180°-30°-70° = 80°

If AB = BC, then ∠BCA = ∠BAC= 80° (Angles opposite to equal sides in a triangle are equal)

Now, ∠ECD = ∠BCD – ∠BCA = 80° – 30P = 50° (∵ ∠BCD = 80° and ∠BCA =30°)

Hence, ∠BCD = 80°

and ∠ECD = 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.**

**Solution:**

**Given:** Diagonals NP and QM of a cyclic quadrilateral are diameters of the circle through the vertices M, P, Q and N of the quadrilateral NQPM.

**To prove:** Quadrilateral NQPM is a rectangle.

**Proof:** ∵ ON = OP = OQ = OM (Radii of circle)

Now, ON = OP = 12 NP

and OM = OQ = 12 MQ

∴ NP = MQ

Hence, the diagonals of the quadrilateral MPQN are equal and bisect each other. So, quadrilateral NQPM is a rectangle.

**Question 8.**

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

**Solution:**

**Given:** Non-parallel sides PS and QR of a trapezium PQRS are equal.

**To prove:** ABCD is a cyclic trapezium.

**Construction:** Draw SM ⊥ PQ and RN ⊥ PQ.

Proof In ∆SMP and ∆RNQ, we get

SP = RQ (Given)

∠SMP = ∠RNQ (Each = 90°)

and SM = RN

(∵ Distance between two parallel lines is always equal)

∴ By RHS criterion, we get

∆ SMP ≅ ∆ RNQ

So, ∠P = ∠Q (By CPCT)

and ∠PSM = ∠QRN

Now, ∠PSM = ∠QRN

∴ 90° + ∠PSM = 90° + ∠QRN (Adding both sides 90°)

∴ ∠MSR + ∠PSM = ∠NRS + ∠QRN (∵∠MSR = ∠NRS = 90°)

So, ∠PSR = ∠QRS

i.e., ∠S = ∠R

Thus, ∠P = ∠Q and ∠R = ∠S …(i)

∴ ∠P+ ∠Q+ ∠R+ ∠S = 360° (∵ Sum of the angles of a quadrilateral is 360°)

∴ 2∠S + ∠Q = 360° [From Eq. (i)]

∠S+∠O = 180°

Hence, PQRS is a cyclic trape∠ium.

**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.**

**Solution:**

**Given:** 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.

**To prove:** ∠ACP = ∠QCD

**Proof:** In circle I, ∠ACP = ∠ABP (Angles in the same segment) …(i)

In circle II, ∠QCD = ∠QBD{Angles in the same segment)…(ii)

∠ABP = ∠QBD (Vertically opposite angles)

From Eqs. (i) and (ii), we get ∠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.**

**Solution:**

**Given:** Two circles are drawn with sides AC and AB of AABC as diameters . Both circles intersect each other at D.

**To prove:** D lies on BC.

**Construction:** Join AD.

**Proof:** Since, AC and AB are the diameters of the two circles.

∠ADB = 90° ( ∴ Angles in a semi-circle) …(i)

and ∠ADC = 90° (Angles in a semi-circle) …(ii)

On adding Eqs. (i) and (ii), we get

∠ADB + ∠ADC = 90° + 90° = 180°

Hence, BCD is a straight line.

So, D lies on BC.

**Question 11.**

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

**Solution:**

Since, ∆ADC and ∆ABC are right angled triangles with common hypotenuse.

Draw a circle with AC as diameter passing through B and D. Join BD.

∵ Angles in the same segment are equal.

∴ ∠CBD = ∠CAD

**Question 12.**

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

**Solution:**

**Given:** PQRS is a parallelogram inscribed in a circle.

**To prove**: PQRS is a rectangle.

**Proof:** Since, PQRS is a cyclic quadrilateral.

∴ ∠P+∠R = 180°

(∵ Sum of opposite angles in a cyclic quadrilateral is 180°) …(i)

But ∠P = ∠R (∵ In a || gm opposite angles are equal) …(ii)

From Eqs. (i) and (ii), we get

∠P = ∠R = 90°

Similarly, ∠Q = ∠S = 90

∴ Each angle of PQRS is 90°.

Hence, PQRS is a rectangle.

**NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.6**

**Question 1.**

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

**Solution:**

**Given:** Two circles with centres O and O’ which intersect each other at C and D.

**To prove:** ∠OCO’ = ∠ODO’

**Construction:** Join OC, OD, O’C and O’D

**Proof:** In ∆ OCO’and ∆ODO’, we have

OC = OD (Radii of the same circle)

O’C = O’D (Radii of the same circle)

OO’ = OO’ (Common)

∴ By SSS criterion, we get

∆ OCO’ ≅ ∆ ODO’

Hence, ∠OCO’ = ∠ODO’ (By CPCT)

**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.**

**Solution:**

Let O be the centre of the given circle and let its radius be cm.

Draw ON ⊥ AB and OM⊥ CD since, ON ⊥ AB, OM ⊥ CD and AB || CD, therefore points N, O, M are collinear.

Let ON = a cm

∴ OM = (6 – a) cm

Join OA and OC.

Then, OA = OC = b c m

Since, the perpendicular from the centre to a chord of the circle bisects the chord.

Therefore, AN = NB= 2.5 cm and OM = MD = 5.5 cm

In ∆OAN and ∆OCM, we get

OA^{2} = ON^{2} + AN^{2}

OC^{2} = OM^{2} + CM^{2}

⇒ b^{2} = a^{2} + (2.5)^{2}

and, b^{2} = (6-a)^{2} + (5.5)^{2} …(i)

So, a^{2} + (2.5)^{2} = (6 – a)^{2} + (5.5)^{2}

⇒ a^{2} + 6.25= 36-12a + a^{2} + 30.25

⇒ 12a = 60

⇒ a = 5

On putting a = 5 in Eq. (i), we get

b^{2} = (5)^{2} + (2.5)^{2}

= 25 + 6.25 = 31.25

So, r = 31.25−−−−√ = 5.6cm (Approx.)

**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 ?**

**Solution:**

Let PQ and RS be two parallel chords of a circle with centre O such that PQ = 6 cm and RS = 8 cm.

Let a be the radius of circle.

Draw ON ⊥ RS, OM ⊥ PQ. Since, PQ || RS and ON ⊥ RS, OM⊥ PQ, therefore points 0,N,M are collinear.

∵ OM = 4 cm and M and N are the mid-points of PQ and RS respectively.

PM = MQ = 12 PQ = 62 = 3 cm

and RN = NS = 12 RS = 82 = 4 cm

In ∆OPM, we have

OP^{2} = OM^{2} + PM^{2}

⇒ a^{2} =4^{2} + 3^{2} = 16 + 9 = 25

⇒ a = 5

In ∆ORN, we have

⇒ OR^{2} = ON^{2} + RN^{2}

⇒ a^{2} = ON^{2} + (4)^{2}

⇒ 25 = ON^{2} + 16

⇒ ON^{2} = 9

⇒ ON = 3cm

Hence, the distance of the chord PS from the centre is 3 cm.

**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.**

**Solution:**

Since, an exterior angle of a triangle is equal to the sum of the interior opposite angles.

∴ In ∆BDC, we get

∠ADC = ∠DBC + ∠DCB …(i)

Since, angle at the centre is twice at a point on the remaining part of circle.

∴ ∠DCE = 12 ∠DOE

⇒ ∠DCB = 12 ∠DOE (∵ ∠DCE = ∠DCB)

∠ADC = 12 ∠AOC

∴ 12 ∠AOC = ∠ABC + 12 ∠DOE (∵ ∠DBC = ∠ABC)

∴ ∠ABC = 12 (∠AOC – ∠DOE)

Hence, ∠ABC is equal to half the difference of angles subtended by the chords AC and DE 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.**

**Solution:**

**Given:** PQRS is a rhombus. PR and SQ are its two diagonals which bisect each other at right angles.

**To prove:** A circle drawn on PQ as diameter will pass through O.

**Construction:** Through O, draw MN || PS and EF || PQ.

**Proof :** ∵ PQ = SR ⇒ 12 PQ = 12 SR

So, PN = SM

Similarly, PE = ON

So, PN = ON = NQ

Therefore, a circle drawn with N as centre and radius PN passes through P, O, Q.

**Question 6.**

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

**Solution:**

Since, ABCE is a cyclic quadrilateral, therefore

∠AED+ ∠ABC= 180°

(∵ Sum of opposite angle of a cyclic quadrilateral is 180°) .. .(i)

∵ ∠ADE + ∠ADC = 180° (EDC is a straight line)

So, ∠ADE + ∠ABC = 180°

(∵ ∠ADC = ∠ABC opposite angle of a || gm).. .(ii)

From Eqs. (i) and (ii), we get

∠AED + ∠ABC = ∠ADE + ∠ABC

⇒ ∠AED = ∠ADE

∴ In ∆AED We have

∠AED = ∠ADE

So, AD = AE

(∵ Sides opposite to equal angles of a triangle are equal)

**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.**

**Solution:**

(i) Let BD and AC be two chords of a circle bisect at P.

In ∆APB and ∆CPD, we get

PA = PC ( ∵ P is the mid-point of AC)

∠APB = ∠CPD (Vertically opposite angles)

and PB = PD (∵ P is the mid-point of BD)

∴ By SAS criterion

∆CPD ≅ ∆APB

∴ CD= AB (By CPCT) …(i)

∴ BD divides the circle into two equal parts. So, BD is a diameter.

Similarly, AC is a diameter.

(ii) Now, BD and AC bisect each other.

So, ABCD is a parallelogram.

Also, AC = BD

∴ 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° – **12** A, 90° – **12** B and 90° – **12** C.**

**Solution:**

∵ ∠EDF = ∠EDA + ∠ADF

∵ ∠EDA and ∠EBA are the angles in the same segment of the circle.

∴ ∠EDA = ∠EBA

and similarly ∠ADF and ∠FCA are the angles in the same segment and hence

**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.**

**Solution:**

Let O’ and O be the centres of two congruent circles.

Since, AB is a common chord of these circles.

∴ ∠BPA = ∠BQA

(∵ Angle subtended by equal chords are 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.**

**Solution:**

(i) Let bisector of ∠A meet the circumcircle of ∆ABC at M.

Join BM and CM.

∴ ∠MBC = ∠MAC (Angles in same segment)

and ∠BCM = ∠BAM (Angles in same segment)

But ∠BAM = ∠CAM (∵ AM is bisector of ∠A)…. .(i)

∴ ∠MBC = ∠BCM

So, MB = MC (Sides opposite to equal angles are equal)

So, M must lie on the perpendicular bisector of BC

(ii) Let M be a point on the perpendicular bisector of BC which lies on circumcircle of ∆ ABC.

Join AM.

Since, M lies on perpendicular bisector of BC.

∴ BM = CM

∠MBC = ∠MCB

But ∠MBC = ∠MAC (Angles in same segment)

and ∠MCB = ∠BAM (Angles in same segment)

So, from Eq. (i),

∠BAM = ∠CAM

AM is the bisector of A.

Hence, bisector of ∠A and perpendicular bisector of BC at M which lies on circumcircle of ∆ABC.