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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°
... AO2 = AP2 + OP2

          = 122 + OP2
 ⇒ 132 = 122 + OP2
 ⇒ 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°

OB2 = OE2 + EB2 = x2 + 25          .................... (i)

and OD2 = OF2 + FD2 = (17 - x)2 + 144   .................... (ii)

But OB = OD
OB2 = OD2

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

x2 + 25 = (17 - x)2 + 144
x = 12       

Substituting x = 12 in Equation (i).

We get r2 = 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°

OB2 = OD2 + DB2
132 = 52 + DB2DB = 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

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 = BE

Thus, 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°
... OB2 = BD2 + OD2
    25 = x2 + y2                        (i)

Again In
D ABD, ∠ D = 90°
...  AB2 = BD2 + AD2
36 = x2 + (5 - y)2                   (ii)

Using eqn (i) in eqn(ii) we get,
36 = 25 - y2 + (5-y)2
⇒ y = cm.
Putting y =
  cm in Eq. (i)
25 = x2

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 DABC 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 DABC 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 = 2BAC
  Þ 2BOD = 2BAC ⇒ ∠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°
21 + 23 = 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 = 2AEB = 2AFB

..
AOB = AEB + AFB
..a = 4 + 3 = 24   ............. (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 = 24 = 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 = 2ABC    ....... (i)

BOC = 2BAC   ....... (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 = 2ABC + 2BAC
⇒ ∠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:

Given:
 AB is a diameter of a circle and radius OD  AB.
∴ BOD = 90° 
BOD = 2BAD ⇒ ∠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 () = 2BCD
                                               = 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,

OB2 = ON2 + NB2                   (By Pythagoras Theorem) 

r2 = (6 – x)2 +   …..(1)

In right triangle OMD, ( By Pythagoras Theorem)

r2 = x2 +       …..(2)

From (1) and (2), we get

(6 – x)2   = x2   

36 – 12x + x2 + = x2 +

12x = 36 + -

12x = 12

x = = 1


Putting x = 1 in (2), we get

r2 = (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,

   OB2 = OM2 + MB2                          ( 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,

            OD2 = ON2 + ND2                      (By Pythagoras Theorem)

             (5)2 = ON2 + (4)2

             25 = ON2+16

             ON2 = 25-16

             ON2 = 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.
 
 

Proof:
 
AB is the common chord of the two congruent circles
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.




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