In This Post we are providing Chapter 10 Circles NCERT Solutions for Class 9 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Circles Class 9 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.
We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Maths Circles NCERT Written Solutions & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.
Exercise 10.1
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.
Answer
(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.
2. Write True or False: Give reasons 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.
Answer
(i) True.
All the line segment from the centre to the circle is of equal length.
(ii) False.
We can draw infinite numbers of equal chords.
(iii) False.
We get major and minor arcs for unequal arcs. So, for equal arcs on circle we can’t say it is major arc or minor arc.
(iv) True.
A chord which is twice as long as radius must pass through the centre of the circle and is diameter to the circle.
(v) False.
Sector is the region between the arc and the two radii of the circle.
(vi) True.
A circle can be drawn on the plane.
Exercise 10.2
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.
Answer
A circle is a collection of points whose every every point is equidistant from the centre. Thus, two circles can only be congruent when they the distance of every point of the both circle is equal from the centre.
Given,
AB = CD (Equal chords)
To prove,
∠AOB = ∠COD
Proof,
In ΔAOB and ΔCOD,
OA = OC (Radii)
OB = OD (Radii)
AB = CD (Given)
∴ ΔAOB ≅ ΔCOD (SSS congruence condition)
Thus, ∠AOB = ∠COD by CPCT.
Equal chords of congruent circles subtend equal angles at their centres.
2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Answer
Given,
∠AOB = ∠COD (Equal angles)
To prove,
AB = CD
Proof,
In ΔAOB and ΔCOD,
OA = OC (Radii)
∠AOB = ∠COD (Given)
OB = OD (Radii)
∴ ΔAOB ≅ ΔCOD (SAS congruence condition)
Thus, AB = CD by CPCT.
If chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Exercise 10.3
1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Answer
One point P is common.
One point P is common.
Two points P and Q are common.
No point is common.
2. Suppose you are given a circle. Give a construction to find its centre.
Answer
Steps of construction:
Step I: A circle is drawn.
Step II: Two chords AB and CD are drawn.
Step III: Perpendicular bisector of the chords AB and CD are drawn.
Step IV: Let these two perpendicular bisector meet at a point. The point of intersection of these two perpendicular bisector is the centre of the circle.
3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Answer
Given,
Two circles which intersect each other at P and Q.
To prove,
OO’ is perpendicular bisector of PQ.
Proof,
In ΔPOO’ and ΔQOO’,
OP = OQ (Radii)
OO’ = OO’ (Common)
O’P = OQ (Radii)
∴ ΔPOO’ ≅ ΔQOO’ (SSS congruence condition)
Thus,
∠POO’ = ∠QOO’ — (i)
In ΔPOR and ΔQOR,
OP = OQ (Radii)
∠POR = ∠QOR (from i)
OR = OR (Common)
∴ ΔPOR ≅ ΔQOR (SAS congruence condition)
Thus,
∠PRO = ∠QRO
also,
∠PRO + ∠QRO = 180°
⇒ ∠PRO = ∠QRO = 180°/2 = 90°
Hence,
OO’ is perpendicular bisector of PQ.
Exercise 10.4
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.
Answer
OP = 5cm, PS = 3cm and OS = 4cm.
also, PQ = 2PR
Let RS be x.
In ΔPOR,
OP2 = OR2 + PR2
⇒ 52 = (4-x)2 + PR2
⇒ 25 = 16 + x2 – 8x + PR2
⇒ PR2 = 9 – x2 + 8x — (i)
In ΔPRS,
PS2 = PR2 + RS2
⇒ 32 = PR2 + x2
⇒ PR2 = 9 – x2 — (ii)
Equating (i) and (ii),
9 – x2 + 8x = 9 – x2
⇒ 8x = 0
⇒ x = 0
Putting the value of x in (i) we get,
PR2 = 9 – 02
⇒ PR = 3cm
Length of the cord PQ = 2PR = 2×3 = 6cm
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.
Answer
Given,
AB and CD are chords intersecting at E.
AB = CD
To prove,
AE = DE and CE = BE
Construction,
OM ⊥ AB and ON ⊥ CD. OE is joined.
Proof,
OM bisects AB (OM ⊥ AB)
ON bisects CD (ON ⊥ CD)
As AB = CD thus,
AM = ND — (i)
and MB = CN — (ii)
In ΔOME and ΔONE,
∠OME = ∠ONE (Perpendiculars)
OE = OE (Common)
OM = ON (AB = CD and thus equidistant from the centre)
ΔOME ≅ ΔONE by RHS congruence condition.
ME = EN by CPCT — (iii)
From (i) and (ii) we get,
AM + ME = ND + EN
⇒ AE = ED
From (ii) and (iii) we get,
MB – ME = CN – EN
⇒ EB = CE
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.
Answer
Given,
AB and CD are chords intersecting at E.
AB = CD, PQ is the diameter.
To prove,
∠BEQ = ∠CEQ
Construction,
OM ⊥ AB and ON ⊥ CD. OE is joined.
In ΔOEM and ΔOEN,
OM = ON (Equal chords are equidistant from the centre)
OE = OE (Common)
∠OME = ∠ONE (Perpendicular)
ΔOEM ≅ ΔOEN by RHS congruence condition.
Thus,
∠MEO = ∠NEO by CPCT
⇒ ∠BEQ = ∠CEQ
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 (see Fig. 10.25).
Answer
OM ⊥ AD is drawn from O.
OM bisects AD as OM ⊥ AD.
⇒ AM = MD — (i)
also, OM bisects BC as OM ⊥ BC.
⇒ BM = MC — (ii)
From (i) and (ii),
AM – BM = MD – MC
⇒ AB = CD
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 6m each, what is the distance between Reshma and Mandip?
Answer
Let A, B and C represent the positions of Reshma, Salma and Mandip respectively.
AB = 6cm and BC = 6cm.
Radius OA = 5cm
BM ⊥ AC is drawn.
ABC is an isosceles triangle as AB = BC, M is mid-point of AC. BM is perpendicular bisector of AC and thus it passes through the centre of the circle.
Let AM = y and OM = x then BM = (5-x).
Applying Pythagoras theorem in ΔOAM,
OA2 = OM2 + AM2
⇒ 52 = x2 + y2 — (i)
Applying Pythagoras theorem in ΔAMB,
AB2 = BM2 + AM2
⇒ 62 = (5-x)2 + y2 — (ii)
Subtracting (i) from (ii), we get
36 – 25 = (5-x)2 – x2 –
⇒ 11 = 25 – 10x
⇒ 10x = 14 ⇒ x= 7/5
Substituting the value of x in (i), we get
y2 + 49/25 = 25
⇒ y2 = 25 – 49/25
⇒ y2 = (625 – 49)/25
⇒ y2 = 576/25
⇒ y = 24/5
Thus,
AC = 2×AM = 2×y = 2×(24/5) m = 48/5 m = 9.6 m
Distance between Reshma and Mandip is 9.6 m.
6. A circular park of radius 20m 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.
Answer
Let A, B and C represent the positions of Ankur, Syed and David respectively. All three boys at equal distances thus ABC is an equilateral triangle.
AD ⊥ BC is drawn. Now, AD is median of ΔABC and it passes through the centre O.
Also, O is the centroid of the ΔABC. OA is the radius of the triangle.
OA = 2/3 AD
Let the side of a triangle a metres then BD = a/2 m.
Applying Pythagoras theorem in ΔABD,
AB2 = BD2 + AD2
⇒ AD2 = AB2 – BD2
⇒ AD2 = a2 – (a/2)2
⇒ AD2 = 3a2/4
⇒ AD = √3a/2
OA = 2/3 AD ⇒ 20 m = 2/3 × √3a/2
⇒ a = 20√3 m
Length of the string is 20√3 m.
1. In Fig. 10.36, A,B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
Answer
Here,
∠AOC = ∠AOB + ∠BOC
⇒ ∠AOC = 60° + 30°
⇒ ∠AOC = 90°
We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.
∠ADC = 1/2∠AOC = 1/2 × 90° = 45°
Page No: 185
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.
Answer
Given,
AB is equal to the radius of the circle.
In ΔOAB,
OA = OB = AB = radius of the circle.
Thus, ΔOAB is an equilateral triangle.
∠AOC = 60°
also,
∠ACB = 1/2 ∠AOB = 1/2 × 60° = 30°
ACBD is a cyclic quadrilateral,
∠ACB + ∠ADB = 180° (Opposite angles of cyclic quadrilateral)
⇒ ∠ADB = 180° – 30° = 150°
3. In Fig. 10.37, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Answer
Reflex ∠POR = 2 × ∠PQR = 2 × 100° = 200°
∴ ∠POR = 360° – 200° = 160°
In ΔOPR,
OP = OR (radii of the circle)
∠OPR = ∠ORP
Now,
∠OPR + ∠ORP +∠POR = 180° (Sum of the angles in a triangle)
⇒ ∠OPR + ∠OPR + 160° = 180°
⇒ 2∠OPR = 180° – 160°
⇒ ∠OPR = 10°
4. In Fig. 10.38, ∠ABC = 69°, ∠ ACB = 31°, find ∠BDC.
∠BAC = ∠BDC (Angles in the segment of the circle)
In ΔABC,
∠BAC + ∠ABC + ∠ACB = 180° (Sum of the angles in a triangle)
⇒ ∠BAC + 69° + 31° = 180°
⇒ ∠BAC = 180° – 100°
⇒ ∠BAC = 80°
Thus, ∠BDC = 80°
5. In Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠ BAC.
∠BAC = ∠CDE (Angles in the segment of the circle)
In ΔCDE,
∠CEB = ∠CDE + ∠DCE (Exterior angles of the triangle)
⇒ 130° = ∠CDE + 20°
⇒ ∠CDE = 110°
Thus, ∠BAC = 110°
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.
Answer
For chord CD,
∠CBD = ∠CAD (Angles in same segment)
∠CAD = 70°
∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°
∠BCD + ∠BAD = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠BCD + 100° = 180°
⇒ ∠BCD = 80°
In ΔABC
AB = BC (given)
∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)
∠BCA = 30°
also, ∠BCD = 80°
∠BCA + ∠ACD = 80°
⇒ 30° + ∠ACD = 80°
∠ACD = 50°
∠ECD = 50°
7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Answer
Let ABCD be a cyclic quadrilateral and its diagonal AC and BD are the diameters of the circle through the vertices of the quadrilateral.
Thus, ABCD is a rectangle as each internal angle is 90°.
8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Answer
Given,
ABCD is a trapezium where non-parallel sides AD and BC are equal.
Construction,
DM and CN are perpendicular drawn on AB from D and C respectively.
To prove,
ABCD is cyclic trapezium.
In ΔDAM and ΔCBN,
AD = BC (Given)
∠AMD = ∠BNC (Right angles)
DM = CN (Distance between the parallel lines)
ΔDAM ≅ ΔCBN by RHS congruence condition.
Now,
∠A = ∠B by CPCT
also, ∠B + ∠C = 180° (sum of the co-interior angles)
⇒ ∠A + ∠C = 180°
Thus, ABCD is a cyclic quadrilateral as sum of the pair of opposite angles is 180°.
Page No: 186
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 Fig. 10.40). Prove that ∠ACP = ∠QCD.
Chords AP and DQ are joined.
For chord AP,
∠PBA = ∠ACP (Angles in the same segment) — (i)
For chord DQ,
∠DBQ = ∠QCD (Angles in same segment) — (ii)
ABD and PBQ are line segments intersecting at B.
∠PBA = ∠DBQ (Vertically opposite angles) — (iii)
By the equations (i), (ii) and (iii),
∠ACP = ∠QCD
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.
Answer
Given,
∠ADB = ∠ADC = 90° (Angle in the semi circle)
Now,
∠ADB + ∠ADC = 180°
⇒ ∠BDC is straight line.
Thus, D lies on the BC.
11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Answer
Given,
AC is the common hypotenuse. ∠B = ∠D = 90°.
To prove,
∠CAD = ∠CBD
Proof:
Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi circle and AC is the diameter of the circle.
⇒ Points A, B, C and D are concyclic.
Thus, CD is the chord.
⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)
12. Prove that a cyclic parallelogram is a rectangle.
Answer
Given,
ABCD is a cyclic parallelogram.
To prove,
ABCD is rectangle.
∠1 + ∠2 = 180° (Opposite angles of a cyclic parallelogram)
also, Opposite angles of a cyclic parallelogram are equal.
Thus,
∠1 = ∠2
⇒ ∠1 + ∠1 = 180°
⇒ ∠1 = 90°
One of the interior angle of the paralleogram is right angled. Thus, ABCD is a rectangle.
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
OA2 = ON2 + AN2
OC2 = OM2 + CM2
⇒ b2 = a2 + (2.5)2
and, b2 = (6-a)2 + (5.5)2 …(i)
So, a2 + (2.5)2 = (6 – a)2 + (5.5)2
⇒ a2 + 6.25= 36-12a + a2 + 30.25
⇒ 12a = 60
⇒ a = 5
On putting a = 5 in Eq. (i), we get
b2 = (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
OP2 = OM2 + PM2
⇒ a2 =42 + 32 = 16 + 9 = 25
⇒ a = 5
In ∆ORN, we have
⇒ OR2 = ON2 + RN2
⇒ a2 = ON2 + (4)2
⇒ 25 = ON2 + 16
⇒ ON2 = 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.
Ex 10.6 Class 9 Maths 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
Ex 10.6 Class 9 Maths 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
Ex 10.6 Class 9 Maths 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.
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