# Question-1

**The radius of a circle is 13 cm and length of one of its chords is 24 cm. Find the distance of the chord from the centre.**

**Solution:**

Let AB = 24 cm be the chord of the circle with radius AO = 13 cm.

Draw OP ⊥ AB. Join OA according to theorem,

AP= ×24 = 12 cm

In Δ APO, ∠P = 90°

.

^{.}. AO

^{2 }= AP

^{2 }+ OP

^{2}

= 12

^{2 }+ OP

^{2}

⇒ 13

^{2 }= 12

^{2 }+ OP

^{2}

⇒ OP = 5 cm

# Question-2

**AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If chords are on opposite sides of the centre of the circle and distance between them is 17 cm find the radius of the circle.**

**Solution:**

Let AB and CD be two parallel chords of the circle C(O, r). Draw OE ⊥ AB and OF ⊥ CD since AB || CD, hence points E, O and F will be collinear, and EF = 17 cm.

Let OE = x cm, then OF = (17 - x) cm. Join 0B and OD.

This follows OB = OD = r

Again EB = AB = 5 cm

and FD = CD = 12 cm

In ΔOEB, ∠E = 90°

∴OB

^{2 }= OE

^{2 }+ EB

^{2 }= x

^{2 }+ 25 .................... (i)

and OD

^{2 }= OF

^{2 }+ FD

^{2 }= (17 - x)

^{2 }+ 144 .................... (ii)

But OB = OD ⇒ OB

^{2 }= OD

^{2}

From Equations (i) and (ii) we get,

x

^{2 }+ 25 = (17 - x)

^{2 }+ 144 ⇒ x = 12

Substituting x = 12 in Equation (i).

We get r

^{2 }= 169 ⇒ r = 13 cm

**∴**

**The radius of the circle**

**=**13 cm

# Question-3

**Determine the length of a chord, which is at a distance of 5 cm from the centre of the circle of radius 13 cm.**

**Solution:**

Let AB be the chord of a circle of radius 13 cm.

Draw OD ⊥ AB, then OD = 5 cm

In ΔODB, ∠D = 90°

\ OB

^{2 }= OD

^{2 }+ DB

^{2}

⇒ 13

^{2 }= 5

^{2 }+ DB

^{2}⇒ DB = 12 cm

Hence, AB = 2 × DB = 24 cm.

# Question-4

**In the Fig. AB and AC are two equal chords of a circle. Prove that bisector of****∠BAC passes through the centre of the circle.****Solution:**

Let AB and AC be two equal chords of a circle and ∠BAD = ∠CAD in ΔAMB and ΔAMC, AB = AC (given), AM = AM (Common) and

∠BAD = ∠CAD.

.

^{.}. ΔAMB @ DAMC

.

^{.}. BM = CM and ∠CMA = ∠BMA=90°

Hence, AD is a perpendicular bisector of chord BC. But perpendicular bisector of a chord passes through the centre.

Hence AD passes through the centre O.

# Question-5

**In the adjoining figure, OD is perpendicular to the chord AB of a circle whose centre is O. If BC is a diameter, show that AC = 2 OD and AC**

**||**

**OD.**

**Solution:**

AB is the chord and OD ⊥ AB ⇒ BD = AD

In ΔABC, D is the mid-point of AB and O is the mid-point of BC

∴ OD || AC

Now in, ΔABC, OD || AC

\ I By BPT Theorem

or AC = 2DO = 2OD

∴ AC = 2 OD and AC || OD.

# Question-6

**In an equilateral triangle, prove that the centroid and the centre of the circum circle (circumcentre) coincide.**

**Solution:**

An equilateral ΔABC such that AB = BC = AC

BC = BC, ∠B = ∠C = 60°

BF = CE (AB = AC)

.

.

Thus, AD = BE = CF

⇒ CF

⇒ GA = GB = GC (

Also OA = OB = OC, where O is the circum-centre.

Hence, G coincides with O.

**To Prove:**The centroid and circum-centre of ΔABC coincide.**Construction:**Draw three medians AD, BE and CF intersecting at G. Then G is the centroid of ΔABC.**Proof:**In ΔBFC and ΔCEB,BC = BC, ∠B = ∠C = 60°

BF = CE (AB = AC)

.

^{.}. D BFC @ D CEB.

^{.}. BE= CF, similarly, AD = BEThus, AD = BE = CF

⇒ CF

⇒ GA = GB = GC (

^{.}.^{.}the centroid divides each median in the ratio 2 : 1)Also OA = OB = OC, where O is the circum-centre.

Hence, G coincides with O.

# Question-7

**In a circle of radius 5 cm, AB and AC are two chords of 6 cm each. Find the length of the chord BC.**

**Solution:**

Let AB and AC be two equal chords, of 6 cm length, of a circle whose centre is O. Join BC. Draw AD ⊥ BC and Join D to O.

Now AB = AC= 6 cm

AB = AC = 6 cm and AD ⊥ BC

∴ BD = DC

⇒ AD is the perpendicular bisector of BC. Also centre O will also lie on the perpendicular bisector of a chord. Join OB and OD.

Now BD = DC = x (say) and OD = y

In ΔOBD, ∠D = 90°

.

^{.}. OB

^{2 }= BD

^{2 }+ OD

^{2}

25 = x

^{2 }+ y

^{2 }(i)

Again In D ABD, ∠ D = 90°

.

^{.}. AB

^{2 }= BD

^{2 }+ AD

^{2}⇒ 36 = x

^{2 }+ (5 - y)

^{2 }(ii)

Using eqn (i) in eqn(ii) we get,

36 = 25 - y

^{2 }+ (5-y)

^{2 }⇒ y = cm.

Putting y = cm in Eq. (i)

25 = x

^{2 }+

⇒ x = 4.8 cm

BC= 2x = 2 * 4.8 = 9.6 cm

# Question-8

**Prove that the angle bisectors of a cyclic quadrilateral form another cyclic quadrilateral.**

**Solution:**

A cyclic quadrilateral ABCD in which AP, BP, CR and DR are angle bisectors of angles A, B, C, D respectively and form a new quadrilateral PQRS (Fig.)

**To Prove:**PQRS is cyclic

i.e. ∠Q + ∠S = 180° and ∠R + ∠P = 180°

**Proof:**In ΔDRC, ∠DRC + ∠RCD + ∠CDR = 180° ....................(i)

In ΔAPB, ∠APB + ∠PBA + ∠BAP = 180° ....................(ii)

Adding equations (i) and (ii) we get,

∠DRC + ∠RCD + ∠CDR + ∠APB + ∠PBA + ∠BAP = 360° ....................(iii)

⇒ ∠DRC + ∠C + ∠D + ∠APB + ∠B + ∠A = 360°

⇒ (∠DRC + ∠APB)+ (∠A + ∠B + ∠C + ∠D) = 360°

⇒ (∠DRC + ∠APB)+ × 360° = 360°

⇒ ∠DRC + ∠APB = 180°

but ∠DRC = ∠QRS (Vertically opposite angles)

and ∠APB = ∠QPS

∴ ∠QRS + ∠QPS = 180°

Thus PQRS is cyclic, as opposite pairs of angles are supplementary.

# Question-9

**Prove that the altitudes of a triangle are concurrent (meet in one point).**

**Solution:**

ΔABC in which BE and CF are perpendiculars on AC and AB respectively and intersect at O. Join AO and produce it to meet BC on D (fig.)

**To Prove:**∠ADC = 90°

**Construction:**Join EF.

**Proof:**∠BEC = ∠BFC = 90°

Points B, C, E, F are concylic ( ∵ BE and CF are perpendiculars on AC and AB respectively and intersect at O)

∠ACD + ∠BFE = 180°

⇒ ∠ACD + 90° + ∠OFE = 180°

⇒ ∠ACD + ∠OFE = 90° .................... (i)

Now ∠OFA + ∠OEA = 180° .................... (ii)

Hence, A, F, O, E are concylic.

∠OFE = ∠OAE (angles in the same segment of circle)

From Equations (i) and (ii) we get

∠ACD + ∠OAE = 90°

∠ACD + ∠DAC = 90°

But ∠ADC + (∠ACD + ∠DAC) = 180°

⇒ ∠ADC + 90° = 180°

⇒ ADC = 90°

# Question-10

**Prove that a circle drawn on one of the equal sides of an isosceles triangle as diameter bisects the base of the triangle.****Solution:**

In the figure ΔABC is an isosceles such that AB = BC. Circle C(O,r) is drawn on the side AB as diameter and intersecting side AC in D.

**To Prove:**AD = CD

**Construction:**Join B to D

**Proof:**∠ADB = 90° (angle in a semi-circle)

Now in ΔABD and ΔBDC

AB = BC (given)

BD = BD (common)

∠ADB = ∠BDC (each 90°)

∴ ΔABD ≅ ΔBDC

∴ AD = CD

# Question-11

**Two circles are drawn on the sides AB and AC of**

**D**

**ABC as diameters the circles intersect each other at a point D.**

**Solution:**

In ΔABC,

AB is the diameter of the circle C(O, r) and AC is the diameter of the circle C'(O', r'). The circles intersect at D.

**To Prove:**

∴D lies on BC.

**Construction:**Join A to D.

**Proof:**Angle in a semi-circle is 90°.

∴ ∠ADB = 90° and ∠ADC = 90°

Hence, ∠ADB + ∠ADC=180°

⇒ ∠BDC = 180°

Therefore, B, D, C are collinear i.e. D lies on BC.

# Question-12

**In the figure,AB is a diameter of the circle C(O,r) chord CD is equal to the radius OC. AC and BD when produced meet at P. Prove that****∠APB = 60°.****Solution:**

**Given:**Circle C(O,r) with diameter AB and chord CD, such that OA = OC = CD.

AC and BD when produced meet at P.

**To Prove:**∠APB = 60°

**Construction:**Join OD, AD.

**Proof:**In ΔOCD,

OC = OD = CD ∴ ∠COD = 60°(Angle of a equilateral triangle)

As ∠CAD = ∠COD

=× 60° = 30°

Now, ∠ADP = ∠ADB = 90° (Angle in a semi-circle)

Now in ΔADP, ∠ADP + ∠DPA + ∠DAP = 180° ⇒ 90° + ∠DPA + 30° = 180°

Hence ∠DPA = 60° ⇒ ∠APB = 60°

# Question-13

**Prove that the bisectors of the angles, formed by producing opposite sides (which are not parallel) of a cyclic quadrilateral, intersect each other at right-angle.**

**Solution:**

Let P be a point on a circumcircle of ΔABC. Perpendiculars PL, PM and PN are drawn on the lines through the line segments BC, CA and AB respectively.

ABCD is a cyclic quadrilateral (Fig.) AD and BC when produced meet at P. DC and AB, when produced, meet at O. Angle bisectors of ∠P and ∠O meet at R.

**To Prove:**∠PRO = 90°

**Construction:**Produce PR to meet AB in S.

**Proof:**In ΔPDL and ΔPSB,

∠DPL = ∠BPS (given)

∠PDL = ∠PBS (exterior angle of a cyclic quadrilateral)

.

^{.}. ∠PLD = ∠PSB

But ∠PLD = ∠OLR (Vertically Opposite angles)

⇒ ∠OLR = ∠PSB = ∠RSO

Now in ΔOLR and ΔOSR

∠ROL = ∠ROS (given)

∠RLO = ∠RSO (proved)

∴ ∠ORL = ∠ORS

But ∠ORL + ∠ORS = 180°

∴ ∠ORL = 90°

⇒ ∠PRO = 90°

# Question-14

**Let P be a point on the circumcircle of**

**D**

**ABC and perpendiculars PL, PM and PN are drawn on BC, CA and AB respectively. Show that L, M and N are collinear (lie on the same straight line).**

**Solution:**

**To Prove:**N, M and L are collinear.

**Construction:**Join PA and PC.

**Proof:**∠PMA + ∠ANP = 90° + 90° = 180°

Therefore, points A, M, P, N are concyclic.

.

^{.}. ∠PMN = ∠PAN ................... (i)

(angles in the same segment)

Again ∠PMC = ∠PLC = 90°

Therefore, quadrilateral PMLC is cyclic.

.

^{.}. ∠PML + ∠PCL=180° ................... (ii)

Now, ∠PAB +∠PCB = 180°

and ∠PAN + ∠PAB = 180°

.

^{.}. ∠PAN = ∠PCB = ∠PCL ................... (iii)

From Equations (i) (ii) and (iii) we get,

∠PML + ∠PCL= ∠PML + ∠PAN

= ∠PML + ∠PMN = 180°

Hence, N, M, L are collinear. (NML line is called Simpson's line).

# Question-15

**In the figure, there are two circles intersecting each other at A and B. Prove that the line joining their centres bisects the common chord.**

**Solution:**

Let C(O, r) and C'(O', r) be two circles having the common chord AB.

Join OA, OB, O'A and O'B

In ΔAOO' and ΔOBO',

OA = O'A, OB = O'B and OO' = OO'

.

^{.}. ΔOAO' @ DOBO' .

^{.}. ∠AOP = ∠POB

Now in ΔAOP and ΔBOP,

OA = OB,OP = OP

and ∠AOP = ∠BOP .

^{.}. ΔAOP ≅ ΔBOP

.

^{.}. AP = BP.

# Question-16

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

**Solution:**

**Given:**ABCD is a parallelogram and circle C(O, r) intersects CD at E.

**To Prove:**AE = AD

**Construction:**Join A to E

**Proof:**

In parallelogram ABCD, ∠B = ∠D (opposite angles of parallelogram) and

∠AED = ∠B (Exterior angle of a cyclic quadrilateral is equal to the opposite interior angle)

∴ ∠ AED = ∠ ADE

∴ AE = AD (sides opposite to equal angles in a triangle).

# Question-17

**An equilateral triangle ABC is inscribed in a circle. P is any point on the minor arc BC. Prove that PA = PB + PC.**

**Solution:**

Equilateral ΔABC inscribed in a circle C(O,r) with point P on minor arc BC.

**To Prove:**PA = PB + PC

**Construction:**Produce BP to Q such that PQ = PC. Join C to Q.

**Proof:**ABPC is a cyclic quadrilateral .

^{.}. ∠A + ∠BPC = 180° .............. (i)

But ∠A = 60° [angle of a Equilateral triangle]

.

^{.}. ∠BPC = 180° - 60° = 120°

.

^{.}. ∠CPQ = 180° - ∠BPC

= 180° - 120°

= 60° .............. (ii)

Now in ΔPQC,

∠3 + ∠4 + ∠Q = 180°

.

^{.}. ∠3 + ∠Q = 120°

But ∠3 = ∠Q

.

^{.}. ∠3 = ∠Q = 60°

Hence, ΔPQC is equilateral

∴ ∠2 = ∠3 = 60°

∴ ∠2 + ∠BCP = ∠3 + ∠BCP

⇒ ∠ACP = ∠BCQ

Now in ΔACP and ΔBCQ, ∠1 = ∠5 [Angles in the same segment of the circle]

∠ACP = ∠BCQ [proved above]

and PC = QC [by construction]

.

^{.}. ΔACP @ DBCQ

.

^{.}. AP = BQ = BP + PQ

⇒ AP = BP + PC [PC = PQ].

# Question-18

**ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at right angle at M. Prove that any line passing through M and bisecting any side of the quadrilateral is perpendicular to the opposite side.**

**Solution:**

Let ABCD be a cyclic quadrilateral whose diagonals AC and BD intersect at M at right-angle. MP, bisecting side CD when produced, meets AB in L.

**To Prove:**∠PLB = 90°

**Proof:**∠AMB = ∠BMC (given) ∠CMD = ∠DMA = 90°

DP = CP

∴ PM = PC = PD [

^{.}.

^{.}the median bisecting the hypotenuse of a right-triangle is half of the hypotenuse.]

Now in ΔCPM, MP = CP ⇒ ∠1 = ∠2

But ∠1 = ∠3 [Angles in the same segment of a circle]

and ∠2 = ∠4 (Vertically opposite angles)

.

^{.}. ∠3 = ∠4

But ∠4 + ∠5 = 90° (given)

⇒ ∠3 + ∠5 = 90°

Now in ΔMBL, ∠MBL + ∠BML + ∠MLB = 180°

⇒ ∠3 + ∠5 + ∠MLB = 180°

or 90° + ∠MLB = 180°

Hence, ∠MLB = 90° ⇒∠PLB = 90°

# Question-19

**O is the circum-centre of the triangle ABC and OD****⊥ BC. Prove that ∠BOD = ∠BAC****Solution:**

**Given:**In ΔABC, O is the circumcentre and OD ⊥ BC.

**Construction:**Join BO and CO.

**Proof:**ΔBOD @ DDOC (R.H.S.) ∠BOD = ∠DOC = ∠BOC

Also ∠BOC = 2∠BAC

Þ 2∠BOD = 2∠BAC ⇒ ∠BOD = ∠BAC

# Question-20

**A trapezium will be cyclic if its non-parallel sides are equal.**

**Solution:**

**Given:**ABCD is a trapezium in which AB || CD and AD = BC

**To Prove:**ABCD is cyclic

i.e. ∠B + ∠D = ∠A + ∠C = 180°

**Proof:**In right-triangles ΔADE and ΔBCF, AD = BC and AE = BF.......(Given)

.

^{.}. ΔADE ≅ ΔBCF

.

^{.}. ∠1 = ∠2 and ∠D = ∠C ...... (C.P.C.T.)

⇒ ∠1 + 90° = ∠2 + 90°

⇒ ∠DAB = ∠ABC ................. (i)

Now ∠BAD + ∠ADC = 180° ........ (ii) [Interior angles of the parellel sides]

∠ABC + ∠ADC = 180° [From Equations (i) and (ii)]

Hence ABCD is cyclic.

# Question-21

**The bisectors of the opposite angles of a cyclic quadrilateral intersect the corresponding circle at the points A and B respectively. Prove that AB is the diameter of the circle.**

**Solution:**

PQRS is a cyclic quadrilateral PA and PB are bisectors of PQRS is a cyclic quadrilateral PA and PB are bisectors of ∠SPQ and ∠QRS respectively.

**To Prove:**AB is diameter of the circle.

**Construction:**Join BP and BQ

**Proof:**∠2 = ∠3 (Angles in the same segment of the circle are equal) ............... (i)

As PQRS is cyclic.

∴ ∠SPQ + ∠QRS = 180°

⇒ 2∠1 + 2∠3 = 180°

⇒ ∠1 + ∠3 = 90° ............... (ii)

⇒ ∠1 + ∠2 = 90° (from equations (i) and (ii))

Hence, ∠APB = 90°

**∴**AB is a diameter (

**∵**Angle in a semicircle is 90°)

# Question-22

**In the Fig, O is the centre of the circle, prove that,**∠

**a =**∠

**b +**∠

**c.**

**Solution:**

∠AOB = 2∠AEB = 2∠AFB

.

^{.}. ∠AOB = ∠AEB + ∠AFB

.

^{.}. ∠a = ∠4 + ∠3 = 2∠4 ............. (i)

Now ∠1 = ∠2 (In same segment) ............. (ii)

∠b = ∠4 + ∠1 (Exterior angles equal to sum of two equal interior opposite angles)

⇒ ∠b = ∠4+∠2 ............. (iii) [from (ii)]

∠c = ∠4 - ∠2 ............. (iv) (Since ∠c +∠2 = ∠4)

Adding equations (iii) and (iv) we get,

∠b + ∠c = 2∠4 = ∠a [from(i)]

or ∠a = ∠b + ∠c

# Question-23

**In the figure, P is a point on the chord BC such that AB = AP. Prove that, CP = CQ.**

**Solution:**

Given P is a point on the chord BC and AB = AP

.

^{.}. ∠ABP = ∠APB ...........(i)

∠ABP = ∠AQC ..............(ii)[angle in the same segment of a circle] and

∠APB = ∠CPQ...............(iii) [Vertically opposite angles]

.

^{.}. ∠CPQ = ∠ AQC = ∠PQC [from (i)]

⇒ CQ = PC

or CP = CQ

# Question-24

**Solution:**

AB is a chord of the circle C(O, r)

∠AOC = 2∠ABC ....... (i)

∠BOC = 2∠BAC ....... (ii) (

**∵**Angles at the centre formed by an arc is twice the angle made by the same arc

in the alternate segment)

Adding equations (i) and (ii), we get

∠AOC + ∠BOC = 2∠ABC + 2∠BAC

⇒ ∠AOB = 2(∠CAB + ∠CBA)

# Question-25

**AB is a diameter of a circle C(O,r) and radius OD****⊥ AB. If C is any point on arc DB, then find ∠BAD and ∠CAD.****If ∠BAC = 20**°**, find ∠DAC.****Solution:**

**AB is a diameter of a circle and radius OD ⊥ AB.**

Given:

Given:

∴ ∠BOD = 90°

∠BOD = 2∠BAD ⇒ ∠BAD = ´ 90° = 45°

In ΔABD, ∠ADB = 90° [In semi-circle] ∴ ∠ABD = 90° - ∠BAD = 90° - 45° = 45°

Now ∠ABD = ∠ACD ⇒ 45° = ∠ACD

or ∠ACD = 45°

Again, ∠BDC = ∠BAC [Angles in the same segment] ∴ ∠BAC = 20° [

^{.}.

^{. }∠BDC = 20°]

Now ∠BAC + ∠CAD = 45°

⇒ 20° + ∠CAD = 45°

or ∠CAD = 25°

# Question-26

**AB and AC are two chords of a circle and the diameter passing through 'A' bisects the angle**∠

**BAC. Prove that AB = AC.**

**Solution:**

**Given:**AD is angle bisector of ∠A,

**Construction:**Join BD and DC.

**Proof:**In ΔABD and ΔACD,

AD = AD

∠DAC = ∠DAB (AD is angle bisector of ∠A)

∠ACD = ∠ABD = 90° [Angle in semi-circle is 90°]

**∴**∠ABD ≅ ΔACD

⇒AB = AC (C.P.C.T.E.)

# Question-27

**In the figure find value of 'x'.****Solution:**

From the given figure,

∠ACB = ∠AOB

= ×120 = 60°

Now ∠ACB + ∠BCD = 180°

Þ 60° + ∠BCD = 180°

Þ ÐBCD = 120°

Again in the bigger circle, m () = 2∠BCD

= 2 ´ 120= 240°

Now m ( ) + m() = 360°

⇒ 240° + x° = 360° ⇒ x = 360° - 240°

Þ x = 120°

# Question-28

**Prove that the line of centers of two intersecting circles subtends equal Angles at the two points of intersection.**

**Solution:**

**Given:**Two intersecting circles with centres A and B. Their points of intersection are P and Q.

**To Prove:**∠ APB = ∠ AQB.

**Proof: ** In Δ APB and Δ AQB,

AP = AQ | Radii of a circle

BP = BQ | Radii of a circle

AB = AB | Common

∴ Δ APB ≅ Δ AQB |SSS Rule

∴ ∠ APB = ∠ AQB | CPCT

# Question-29

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

**Solution:**

Let the radius of the circle be r cm. Let OM = x cm.

Then ON = (6 - x) cm.

Given, OM ⊥ CD

∴ M is the mid-point of CD. (

**∵**The perpendicular from the centre of a circle to a chord bisects the chord)

∴ MD = MC = CD =(11)cm= cm

Given, ON ⊥ AB

∴ N is the mid-point of AB (

**∵**The perpendicular from the centre of a circle to a chord bisects the chord)

∴ NB = AN = AB = (5) = cm

In right triangle ONB,

OB

^{2}= ON

^{2}+ NB

^{2 }(By Pythagoras Theorem)

⇒ r

^{2}= (6 – x)

^{2}+ …..(1)

In right triangle OMD, (

^{ }By Pythagoras Theorem)

⇒ r

^{2}= x

^{2}+ …..(2)

From (1) and (2), we get

(6 – x)

^{2}+ = x

^{2}+

⇒ 36 – 12x + x

^{2}+ = x

^{2}+

⇒ 12x = 36 + -

⇒ 12x = 12

⇒ x = = 1

Putting x = 1 in (2), we get

r

^{2}= (1)

^{2}+

= 1 + =

⇒ r =

Hence the radius of the circle is cm.

# Question-30

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

**Solution:**

**Case I. When the two chords are on the same side of the centre**

Given: OM ⊥ AB

∴ M is the mid-point of AB. (

**∵**The perpendicular from the centre of a circle to a chord bisects the chord)

BM = AM = AB = (6) = 3 cm

Given: ON ⊥ CD

∴ N is the mid-point of CD. (

**∵**The perpendicular from the centre of a circle to a chord bisects the chord)

∴ DN = CN = CD = (8) = 4 cm

In right triangle OMB,

OB

^{2}= OM

^{2}+ MB

^{2 }( By Pythagoras Theorem )

= (4)

^{2}+ (3)

^{2}

= 16 + 9 = 25

⇒ OB = = 5 cm

∴ OD = OB = 5 cm ( Radii of a circle)

In right triangle OND,

OD

^{2}= ON

^{2}+ ND

^{2}(By Pythagoras Theorem)

⇒ (5)

^{2}= ON

^{2}+ (4)

^{2}

⇒ 25 = ON

^{2}+16

⇒ ON

^{2}= 25-16

⇒ ON

^{2}= 9

⇒ ON = = 3 cm

Hence the distance of the other chord from the centre is 3 cm.

**Case II. When the two chords are on the opposite sides of the centre**

** **

As in case I

ON = 3 cm.

# Question-31

**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 center.**

**Solution:**

Let ∠ ABC = x, ∠ AOC = y and ∠ DOE = z. …. (1)

**To prove:**x =

∠ C’OD + ∠ A’OE = z - y

Let ∠ C'OD = θ

Then ∠ A'OE = z - y - θ ...(1) (∵ From(1))

∠ AOD = π -(∠ AOC + ∠ C'OD) = π -(y + θ )

∠ COE = π -(∠ C'OA' + ∠ A'OE)

= π - (y + z – y - θ )

= π -(z - θ )

∴ AD = CE

∴ ∠ AOD = ∠ COE (∵ Equal chords subtend equal angles at the centre)

∴ π -(y + θ ) = π -(z - θ )

⇒ y + θ = z - θ

⇒ 2θ = z - y

⇒ θ = ∴ ∠ C'OD =

and ∠ A’OE = z - y - =

∴ ∠ AOD = π - (y + θ )

= π -

= ∠ COE

In Δ OAD,

∴ OA = OD (∵ Radii of the same circle)

∴ ∠ OAD = ∠ ODA (∵ Angles opposite to the same sides of a triangle are equal)

In ΔOAD,

∠ OAD + ∠ ODA + ∠ AOD = π (∵ Sum of all the angles of a triangle is π radians)

⇒ ∠ OAD + ∠ OAD + π - = π

⇒ 2∠ 0AD =

⇒ ∠ 0AD =

Similarly, ∠ OCE =

∴ ∠ OAB = π -

and ∠ OCB = π -

In quadrilateral AOCB,

∠ ABC + ∠ OAB + ∠ OCB + ∠ AOC = 2π (∵ Sum of all the angles of a quadrilateral is 2π radians)

⇒ x + π - + π - + y = 2π

⇒ x + y =

⇒ 2x + 2y = y + z

⇒ 2x = z – y

⇒ x =

Hence the result.

# Question-32

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

**Solution:**

We know that the diagonals of a rhombus bisect each other at right angles.

The circle drawn with one side AB as diameter will pass through the mid-point E of BD which is the point of intersection of the diagonals.

# Question-33

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

**Solution:**

**Given:**ABCD is a parallelogram. The circle through A, B and C intersects CD (produced, if necessary) at E.

**To Prove:**AE = AD

**Proof:**In cyclic quadrilateral ABCE,

∠ AED + ∠ ABC = 180° ...(1) (Since opposite angles of a cyclic quadrilateral are supplementary)

Also, ∠ ADE + ∠ ADC = 180° (Linear Pair Axiom )

But ∠ ADC = ∠ ABC (Opposite angles of a parallelogram )

∴ ∠ ADE +∠ ABC = 180° ...(2)

From (1) and (2), we have

∠ AED + ∠ ABC = ∠ ADE + ∠ ABC

⇒ ∠ AED = ∠ ADE

In triangle ADE,

AE = AD

∴ Sides opposite to equal angles of A are equal. Proved.

# Question-34

**AC and BD are chords of a circle which bisect each other. Prove that AC and BD are diameters and ABCD is a rectangle.**

**Solution:**

**Given:**AC and BD are chords of a circle that bisect each other.

**To Prove:**(i) AC and BD are diameters (ii) ABCD is a rectangle.

**Proof:**(i)Given, AC and BD are chords of a circle which bisect each other

In ΔABD, ∠ A = 90°

∴ BD is a diameter | Since angle in a semi-circle is 90°

In ΔBCD, ∠ D = 90°

∴ AC is a diameter | Single angle in a semi-circle is 90°

Thus, AC and BD are diameters.

(ii) Let the chords AC and BD intersect each other at O. Join AB, BC, CD and DA.

In Δ OAB and Δ OCD,

OA = OC | Given

OB = OD | Given

∠ AOB = ∠ COD | Vertically opposite angles

∴ Δ OAB ≅ Δ OCD | SAS

∴ AB = CD | CPCT

⇒ AB ≅ CD ...(1)

Similarly, we can show that

AD = CB ...(2)

Adding (1) and (2), we get

AB + AD ≅ CD + CB

⇒ BAD = BCD

⇒ BD divides the circle into two equal parts (each a semi-circle) and the angle of a semi-circle is 90°.

∴ ∠ A = 90° and ∠ C = 90°

Similarly, we can show that

∠ B = 90° and ∠ D = 90°

∴ ∠ A = ∠ B = ∠ C = ∠ D = 90°

⇒ ABCD is a rectangle.

# Question-35

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

^{ }– ÐB and 90°–ÐC.**Solution:**

**Given:**Bisectors of angles A, B and C of a triangle ABC intersect its circum circle at D, E and F respectively.

**To Prove:**The angles of the ΔDEF are 90° – , 90° – and 90° – respectively.

**Construction:**Join DE, EF and FD.

**Proof:**∠ FDE = ∠ FDA + ∠ EDA

= ∠ FCA + ∠ EBA | Since angles in the same segment are equal

=

**∠**C +

**∠**B

=∠ D =

**=**| In Δ ABC, ∠ A + ∠ B + ∠ C = 180° (Angle Sum Property)

= 90° –

Similarly, we can show that

∠ E = 90° –

and ∠ F = 90° –

# Question-36

**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:**

**Given:**Two congruent circles intersect each other at points A and B. A line through A meets the circles in P and Q.

**AB is the common chord of the two congruent circles**

Proof:

Proof:

∴ ∠ APB = ∠ AQB | Since angles subtended by equal chords are equal

∴ BP = BQ. | Sides opposite to equal angles are equal

# Question-37

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

**Solution:**

**Given:**Bisector AP of angle A of ΔABC and the perpendicular bisector PQ of its opposite side BC intersect at P.

**To Prove:**P lies on the circumcircle of the triangle ABC.

**Construction:**Draw the circle through three non-collinear points A, B and P.

**Proof:**∠ BAP = ∠ CAP

⇒ chord BP = chord CP

⇒ BP = CP

In Δ BMP and Δ CMP,

BM = CM | M is the bisector of BC

BP = CP |Proved above

MP = MP | Common side

∴ Δ BMP ≅ CMP | SSS

∴ ∠ BMP = ∠CMP | CPCT

But ∠ BMP + ∠ CMP = 180° | Linear Pair Axiom

∴ ∠ BMP = ∠ CMP = 90°

⇒ PM is the right bisector of BC.

**Alieter:**

Assume that C does not lie on the circle though A, B and P. Let this circle intersect the side AC at C’. (say)

∠ APB = ∠ ACB |Given

∠ APB = ∠ AC’B | Angles in the same segment

∴ ∠ ACB = ∠ AC’B

⇒ C and C’ coincide

⇒ The assumption that the point C does not lie on the circle is false.

∴ A, B, P and C are concyclic.