NCERT Solution for Class 9 Mathematics Chapter 10 - Circles RR ACADEMY MEERUT

NCERT Solution for Class 9 Mathematics Chapter 10 - Circles Page/Excercise 10.1

Question 1

Fill in the blanks
(i)  The centre of a circle lies in __________ of the circle. (exterior/interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in       __________ of the circle. (exterior/ interior)
(iii) The longest chord of a circle is a __________ of the circle.
(iv) An arc is a __________ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and __________ of the circle.
(vi) A circle divides the plane, on which it lies, in __________ parts.

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       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.

Question 2

Write True or False: Give reasons for your answers.   (i)   Line segment joining the centre of 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 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.

NCERT Solution for Class 9 Mathematics Chapter 10 - Circles Page/Excercise 10.2

Question 1

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

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.   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) 
 AOB = CO'D            (by CPCT) 
Hence equal chords of congruent circles subtend equal angles at their centres.

Question 2

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

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.

NCERT Solution for Class 9 Mathematics Chapter 10 - Circles Page/Excercise 10.3

Question 1

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

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.                        (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.     

                          
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. 

Question 2

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

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

Question 3

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

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. 

NCERT Solution for Class 9 Mathematics Chapter 10 - Circles Page/Excercise 10.4

Question 1

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

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 
    Given that OO' = 4 cm
    Let OC be x. so, O'C will be 4 - x 
    In OAC
    OA2 = AC2 + OC2  
     52 = AC2 + x2
     25 - x2 = AC2            ... (1) 
    In O'AC
    O'A2 = AC2 + O'C2
     32 = AC2 + (4 - x)2
     9 = AC2 + 16 + x2 - 8x
     AC2 = - x2 - 7 + 8x        ... (2)  
From equations (1) and (2), we have
    25 - x2 = - x2 - 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, AC2 = 25 - x2 = 25 - 42 = 25 - 16 = 9
 AC = 3 m 
The length of the common chord AB = 2 AC = (2  3) m = 6 m 

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 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 90o)
   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. 

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 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 90o)
    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. 

Question 4

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

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

Question 5

Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m 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 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.                                                 AR = AS =  = 3cm    OR = OS = OM = 5 m     (radii of circle)
    In OAR
    OA2 + AR2 = OR2             
    OA2 + (3 m)2 = (5 m)2
    OA2 = (25 - 9) m2 = 16 m2
    OA = 4 m                               
    We know that in an isosceles triangle altitude divides the base, so in RSM
    RCS will be of 90o and RC = CM             
    Area of ORS =   OARS         
                      RC = 4.8     RM = 2RC = 2(4.8)= 9.6        So, 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 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                AB = OA + OB = (20 + 10) m = 30 m. 
    In ABD

    AD2 = AB2 + BD2
    AD2 = (30)2 +    
        
    So, length of string of each phone will be  m. 

NCERT Solution for Class 9 Mathematics Chapter 10 - Circles Page/Excercise 10.5

Question 1

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

Solution 1

We may observe that
    AOC = AOB + BOC
        = 60o + 30o
        = 90o 
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.   

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 2

                              In OAB
    AB = OA = OB = radius
OAB is an equilateral triangle. 
So, each interior angle of this triangle will be of 60o   AOB = 60o   Now,     In cyclic quadrilateral ACBD
ACB + ADB = 180o        (Opposite angle in cyclic quadrilateral)
ADB = 180o - 30o = 150o 
So, angle subtended by this chord at a point on major arc and minor arc are 30o and 150o respectively.

Question 3

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

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 = 180o                    (Opposite angles of cyclic quadrilateral) 
PSR = 180o - 100o = 80o
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 (80o) = 160o In POR
OP = OR                                        (radii of same circle)   OPR = ORP                         (Angles opposite equal sides of a triangle) OPR + ORP + POR = 180o    (Angle sum property of a triangle) 2 OPR + 160o= 180o
2 OPR = 180o - 160o = 20o OPR = 10o 

Question 4

In the given figure, ABC = 69o, ACB = 31o, find BDC. 

Solution 4

In ABC BAC + ABC + ACB = 180o     (Angle sum property of a triangle) 
BAC + 69o + 31o = 180o BAC = 180o - 100º
BAC = 80o BDC = BAC = 80o                    (Angles in same segment of circle are equal) 
    

Question 5

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

Solution 5

In CDE
CDE + DCE = CEB        (Exterior angle)
CDE + 20o = 130o CDE = 110o
But BAC = CDE               (Angles in same segment of circle)
BAC = 110o

Question 6

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

Solution 6

                                 For chord CD CBD = CAD                    (Angles in same segment) CAD = 70o BAD = BAC + CAD = 30o + 70o = 100o 
BCD + BAD = 180o        (Opposite angles of a cyclic quadrilateral)
BCD + 100o = 180o
BCD = 80o 
In ABC AB = BC                               (given)
 BCA = CAB               (Angles opposite to equal sides of a triangle) 
BCA = 30o
We have BCD = 80o
BCA + ACD = 80o 
30o + ACD = 80o ACD = 50o ECD = 50o

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 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 = 180o            (Cyclic quadrilateral) 
    BCD = 180o - 90o = 90o             (Considering AC as a chord)      ADC + ABC = 180o            (Cyclic quadrilateral) 
    90o + ABC = 180o
    ABC = 90o
    Here, each interior angle of cyclic quadrilateral is of 90o. Hence it is a rectangle.

Question 8

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

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 90o)
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 = 180o                ... (2)    
BAD + BCD = 180o          [Using equation (1)] 
This equation shows that the opposite angles are supplementary. 
So, ABCD is a cyclic quadrilateral.

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 the given figure). Prove that ACP = QCD                                          

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 

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 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 = 90o            (Angle subtend by semicircle)
    ADC = 90o            (Angle subtend by semicircle)
    BDC = ADB + ADC = 90o + 90o = 180o 
 Hence BDC is straight line and our assumption was wrong.
 Thus, Point D lies on third side BC of ABC 

                                  

Question 11

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

Solution 11

                        In ABC ABC + BCA + CAB = 180o    (Angle sum property of a triangle) 
 90o + BCA + CAB = 180o
 BCA + CAB = 90o        ... (1) 
In ADC CDA + ACD + DAC = 180o    (Angle sum property of a triangle)
 90o + ACD + DAC = 180o
 ACD + DAC = 90o        ... (2) 
Adding equations (1) and (2), we have BCA + CAB + ACD + DAC = 180o  (BCA + ACD) + (CAB + DAC) = 180o BCD + DAB = 180o        ... (3) 
    But it is given that
B + D = 90o + 90o = 180o        ... (4) 
From equations (3) and (4), we can see that quadrilateral ABCD is having sum of measures of opposite angles as 180o. 
So, it is a cyclic quadrilateral.
Consider chord CD.
Now, CAD = CBD                      (Angles in same segment)                                             

Question 12

Prove that a cyclic parallelogram is a rectangle.

Solution 12

                                          Let ABCD be a cyclic parallelogram.
    A + C = 180o     (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 = 180o
A + A = 180o
2 A = 180o
A = 90o 
Parallelogram ABCD is having its one of interior angles as 90o, so, it is a rectangle.

NCERT Solution for Class 9 Mathematics Chapter 10 - Circles Page/Excercise 10.6

Question 1

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

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',                               Now in AOO'  and BOO' 
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.

Question 2

Two chords AB and CD of lengths 5 cm and 11cm 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 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 NOD    We have OB = OD             (radii of same circle)
So, from equation (1) and (2)    From equation (2)    So, radius of circle is found to be  cm.

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 form the centre?

Solution 3

                                                Distance of smaller chord AB from centre of circle = 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.

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 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) 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) In  ODE,
OD = OE
OED = ODE         (let y)
ADEC is a cyclic quadrilateral 
 CAD + DEC = 180o         (opposite angles are supplementary) x + a + x + y = 180o
2x + a + y = 180o
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 - 360o        ... (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 - 360o]
ABC =  [DOE -  AOC]    [Using equation (5)] 

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 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 90o.
 COD = 90o 
Also in rhombus the diagonals intersect each other at 90o
AOB = BOC = COD = DOA = 90o 
So, point O has to lie on the circle.  

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 6

                                  We see that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral sum of opposite angles is 180o
    AEC + CBA = 180o 
    AEC + AED = 180o        (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) 

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 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)   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 But A + C = 180o      (ABCD is a cyclic quadrilateral)
A + A = 180o
  A = 180o
A = 90o
As ACBD is a parallelogram and one of its interior angles is 90o, so it is a rectangle.
A is the angle subtended by chord BD. And as A = 90o, so BD should be diameter of circle. Similarly AC is diameter of circle.

Question 8

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 90o 

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 

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 9

                                     AB is common chord in both congruent circles.
 APB = AQB   
Now in BPQ 
APB = AQB  
 BP = BQ            (angles opposite to equal sides of a triangle)

Question 10

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

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 OE = OE                                   (common)
OB = OC                                  (radii of same circle) OEB = OEC                       (Each 90o as OD  BC)  BOE  COE                (RHS congruence rule) BOE = COE            (by CPCT)    ... (2) 
But BOE + COE = BOC  BOE +BOE = 2 A        [Using equations (1) and (2)]
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.