NCERT Solutions for Class 9 Maths Chapter 10 Circles

NCERT Solutions for Class 9 Maths Chapter-10 Circles help students master the concepts of chords, tangents, arcs, and angles in circles. This class 9 maths chapter 10 is essential for building a strong foundation in geometry. Using these solutions along with CBSE Class 9 Maths NCERT solutions, class 9 sample papers, and class 9 Maths Chapter 10 solutions helps students build strong problem-solving skills and excel in exams.

NCERT Solutions for Class 9 Maths Chapter 10 Circles

NCERT Solutions for Class 9 Maths Chapter 10 Circles are designed to help students understand every concept in detail. Before attempting exercises, ensure that you have thoroughly read the theory, learned all formulas, and understood the properties of circles. These solutions include tips and shortcuts to solve problems efficiently and are linked with related topics like CBSE Class 9 syllabus and class 9 maths chapter 10 circles

NCERT Solutions for Class 9 Maths Exercise 10.1

This exercise focuses on basic definitions and properties related to circles. Solving these questions strengthens your understanding of the foundational concepts like radius, diameter, chord, and segments. Using NCERT Solutions for Class 9 Maths Exercise 10.1 ensures that students can tackle every problem confidently.

Question 1. Fill in the blanks:

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

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

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

(iv) An arc is a _______________ when its ends are the ends of a diameter.

(v) Segment of a circle is the region between an arc and _______________ of the circle.

(vi) A circle divides the plane, on which it lies, in _______________ parts.

Solution:
(i) Interior

(ii) Exterior

(iii) diameter

(iv) Semi-circle

(v) Chord

(vi) Three

Question 2. Write True or False:

(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, which is twice as long as its radius is a diameter of the circle.

(v) Sector is the region between the chord and its corresponding arc.

(vi) A circle is a plane figure.

Solution:
(i) True

(ii) False

(iii) False

(iv) True

(v) False

(vi) True

NCERT Solutions for Class 9 Maths Exercise 10.2

Exercise 10.2 involves properties of congruent circles and equal chords. These questions test your knowledge of angles and chord relationships. Using NCERT Solutions for Class 9 Maths Exercise 10.2 helps students understand class 9 chapter 10 maths deeply.

Question 1. Recall that two circles are congruent, if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres
Solution:
Given: Two congruent circles with centres O and O’ and radii r, which have chords AB and CD respectively such that AB = CD.
To Prove: ∠AOB = ∠CO’D
Proof: In ∆AOB and ∆CO’D, we have
AB = CD [Given]
OA = O’C [Each equal to r]
OB = O’D [Each equal to r]
∴ ∆AOB ≅ ∆CO’D [By SSS congruence criteria]
⇒ ∠AOB = ∠CO’D [C.P.C.T.]

Question 2. Prove that, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Solution:
Given: Two congruent circles with centres O & O’ and radii r which have chords AB and CD respectively such that ∠AOB = ∠CO’D.

To Prove: AB = CD
Proof: In ∆AOB and ∆CO’D, we have
OA = O’C [Each equal to r]
OB = O’D [Each equal to r]
∠AOB = ∠CO’D [Given]
∴ ∆AOB ≅ ∆CO’D [By SAS congruence criteria]
Hence, AB = CD [C.P.C.T.]

NCERT Solutions for Class 9 Maths Exercise 10.3

Exercise 10.3 includes constructions and understanding of intersecting circles. These questions enhance logical reasoning and geometric drawing skills. Using NCERT Solutions for Class 9 Maths Exercise 10.3 ensures clear understanding of class 9 maths chapter 10.

Question 1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Solution:
Let us draw different pairs of circles as shown below:

We have

NCERT Solutions for Class 9 Maths Chapter-10 Circles

Thus, two circles can have at the most two points in common.

Question 2. Suppose you are given a circle. Give a construction to find its centre.
Solution:
Steps of construction :
Step I : Take any three points on the given circle. Let these points be A, B and C.
Step II : Join AB and BC.
Step III : Draw the perpendicular bisector, PQ of AB.
Step IV: Draw the perpendicular bisector, RS of BC such that it intersects PQ at O.

Thus, ‘O’ is the required centre of the given drcle.

Question 3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Solution:
We have two circles with centres O and O’, intersecting at A and B.
∴ AB is the common chord of two circles and OO’ is the line segment joining their centres.
Let OO’ and AB intersect each other at M.

∴ To prove that OO’ is the perpendicular bisector of AB,
we join OA, OB, O’A and O’B. Now, in ∆QAO’ and ∆OBO’,
we have
OA = OB [Radii of the same circle]
O’A = O’B [Radii of the same circle]
OO’ = OO’ [Common]
∴ ∆OAO’ ≅ ∆OBO’ [By SSS congruence criteria]
⇒ ∠1 = ∠2 , [C.P.C.T.]
Now, in ∆AOM and ∆BOM, we have
OA = OB [Radii of the same circle]
OM = OM [Common]
∠1 = ∠2 [Proved above]
∴ ∆AOM = ∆BOM [By SAS congruence criteria]
⇒ ∠3 = ∠4 [C.P.C.T.]
But ∠3 + ∠4 = 180° [Linear pair]
∴∠3=∠4 = 90°
⇒ AM ⊥ OO’
Also, AM = BM [C.P.C.T.]
⇒ M is the mid-point of AB.
Thus, OO’ is the perpendicular bisector of AB.

NCERT Solutions for Class 9 Maths Exercise 10.4

Exercise 10.4 focuses on chords, segments, and perpendiculars from the centre. Solving these helps in understanding symmetry and distances in circles. The class 9 maths chapter 10 solutions also guide students to approach such problems systematically.

Question 1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centers is 4 cm. Find the length of the common chord.

Solution:
Let two circles with centres O and O’ intersect each other at points A and B. On joining A and B, AB is a common chor

Radius OA = 5 cm, Radius O’A = 3 cm,

Distance between their centers OO’ = 4 cm

In triangle AOO’,

5 ² = 4 ² + 3 ²

25 = 16 + 9

25 = 25

Hence AOO’ is a right triangle, right angled at O’.

Since, perpendicular drawn from the center of the circle bisects the chord.

Hence O’ is the mid-point of the chord AB. Also O’ is the centre of the circle II.

Therefore length of chord AB = Diameter of circle II

Length of chord AB = 2 x 3 = 6 cm.

Question 2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Solution:
Given: Let AB and CD are two equal chords of a circle of centers O intersecting each other at point E within the circle.

To prove: (a) AE = CE

(b) BE = DE

Construction: Draw OM AB, ON CD. Also join OE.

Proof: In right triangles OME and ONE,

OME = ONE =

OM = ON

[Equal chords are equidistance from the centre]

OE = OE [Common]

OME ONE [RHS rule of congruency]

ME = NE [By CPCT] ……….(i)

Now, O is the centre of circle and OM AB

AM = AB

[Perpendicular from the centre bisects the chord] …..(ii)

Similarly, NC = CD ……….(iii)

But AB = CD [Given]

From eq. (ii) and (iii), AM = NC ……….(iv)

Also MB = DN …….…(v)

Adding (i) and (iv), we get,

AM + ME = NC + NE

AE = CE [Proved part (a)]

Now AB = CD [Given]

AE = CE [Proved]

AB – AE = CD – CE

BE = DE [Proved part (b)]

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 chord.

Solution:
Given: AB and CD be two equal chords of a circle with centre O intersecting each other with in the circle at point E. OE is joined.

To prove: OEM = OEN

Construction: Draw OM AB and

ON CD.

Proof: In right angled triangles OME and ONE,

OME = ONE [Each ]

OM = ON [Equal chords are equidistant from the centre]

OE = OE [Common]

OME ONE [RHS rule of congruency]

OEM = OEN [By CPCT]

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. (See figure)

Solution:
Given: Line intersects two concentric circles with centre O at points A, B, C and D.

To prove: AB = CD

Construction: Draw OL

Proof: AD is a chord of outer circle and OL AD.

AL = LD ………(i) [Perpendicular drawn from the centre bisects the chord]

Now,   BC is a chord of inner circle and

OL BC

BL = LC …(ii) [Perpendicular drawn from the centre bisects the chord]

Subtracting (ii) from (i), we get,

AL – BL = LD – LC

AB = CD

Question 5. Three girls Reshma, Salma and Mandip are 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:
Let Reshma, Salma and Mandip takes the position C, A and B on the circle. Since AB = AC

The centre lies on the bisector of BAC.

Let M be the point of intersection of BC and OA.

Again, since AB = AC and AM bisects

CAB.

AM CB and M is the mid-point of CB.

Let OM = then MA =

From right angled triangle OMB,

OB ² = OM ² + MB ²

5 ² = + MB ² ……….(i)

Again, in right angled triangle AMB,

AB ² = AM ² + MB ²

6 ² = + MB ² ……….(ii)

Equating the value of MB2 from eq. (i) and (ii),

Hence, from eq. (i),

MB ² = =

= =

MB = = 4.8 cm

BC = 2MB = 2 x 4.8 = 9.6 cm

Question 6. A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Solution:
Let position of three boys Ankur, Syed and David are denoted by the points A, B and C respectively.

A = B = C = [say]

Since equal sides of equilateral triangle are as equal chords and perpendicular distances of equal chords of a circle are equidistant from the centre.

OD = OE = OF = cm [say]

Join OA, OB and OC.

Area of AOB

= Area of BOC = Area of AOC

And Area of ABC

= Area of AOB + Area of BOC + Area of AOC

And Area of ABC = 3 x Area of BOC

= 3 ( BC x OE)

= 3 ( )

……….(i)

Now, CE BC

BE = EC = BC    [ Perpendicular drawn from the centre bisects the chord]

BE = EC =

BE = EC = [Using eq. (i)]

BE = EC =

Now in right angled triangle BEO,

OE ² + BE ² = OB ² [Using Pythagoras theorem]

= 10 m

And = = m

Thus distance between any two boys is m.

 

NCERT Solutions for Class 9 Maths Exercise 10.5

This exercise deals with angles subtended by arcs, cyclic quadrilaterals, and properties of trapeziums and rectangles. These problems sharpen analytical and visualization skills. Using NCERT Solutions for Class 9 Maths Exercise 10.5 and the ncert class 9 maths chapter 10 circles pdf helps students revise anytime.

Question 1. In figure, A, B, C are three points on a circle with centre O such that BOC = AOB = If D is a point on the circle other than the arc ABC, find ADC.

Solution:
AOC = AOB + BOC

AOC =

Now AOC = 2 ADC

[ Angled subtended by an arc, at the centre of the circle is double the angle subtended by the same arc at any point in the remaining part of the circle]

ADC = AOC

ADC = =

Question 2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord on a point on the minor arc and also at a point on the major arc.

Solution:
Let AB be the minor arc of circle.

Chord AB = Radius OA = Radius OB

AOB is an equilateral triangle.

AOB =

Now

AOB + BOA =

+ BOA =

BOA =

D is a point in the minor arc.

= 2 BDA

BOA = 2 BDA

BDA = BOA =

BDA =

Thus angle subtended by major arc, at any point D in the minor arc is

Let E be a point in the major arc

= 2 AEB

AOB = 2 AEB

AEB = AOB

AEB = =

Question 3. In figure, PQR = where P, Q, R are points on a circle with centre O. Find OPR.

Solution:
In the figure, Q is a point in the minor arc

= 2 PQR

ROP = 2 PQR

ROP = =

Now

POR + ROP =

POR + =

POR = = …..(i)

Now OPR is an isosceles triangle.

OP = OR [radii of the circle]

OPR = ORP [angles opposite to equal sides are equal] …..(ii)

Now in isosceles triangle OPR,

OPR + ORP + POR =

OPR + ORP + =

2 OPR = [Using (i) & (ii)]

2 OPR =

OPR =

Question 4. In figure, ABC = ACB = find BDC.

Solution:
In triangle ABC,

BAC + ABC + ACB =

BAC +

BAC =

BAC = …….(i)

Since, A and D are the points in the same segment of the circle.

BDC = BAC

[Angles subtended by the same arc at any points in the alternate segment of a circle are equal]

BDC = [Using (i)]

Question 5. In figure, A, B, C, D are four points on a circle. AC and BD intersect at a point E such that BEC = and

ECD = Find BAC.

Solution:
Given: BEC = and ECD =

DEC = BEC = [Linear pair]

Now in DEC,

DEC + DCE + EDC = [Angle sum property]

EDC =

EDC =

BAC = EDC = [Angles in same segment]

Question 6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. DBC = BAC is find BCD. Further if AB = BC, find ECD.

Solution:
For chord CD

(Angles in same segment)

=
(Opposite angles of a cyclic quadrilateral)

AB = BC (given)
(Angles opposite to equal sides of a triangle)

We have

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:
Let ABCD a cyclic quadrilateral having diagonals as BD and AC intersecting each other at point O.

(Consider BD as a chord) (Cyclic quadrilateral)

(Considering AC as a chord)

(Cyclic quadrilateral)

Here, each interior angle of cyclic quadrilateral is of . Hence it is a rectangle.

Question 8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution:
Given: A trapezium ABCD in which AB CD and AD = BC.

To prove: The points A, B, C, D are concyclic.

Construction: Draw DE CB.

Proof: Since DE CB and EB DC.

EBCD is a parallelogram.

DE = CB and DEB = DCB

Now AD = BC and DA = DE

DAE = DEB

But DEA + DEB =

DAE + DCB =

[ DEA = DAE and DEB = DCB]

DAB + DCB =

A + C =

Hence, ABCD is a cyclic trapezium.

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, P, Q respectively (see figure). Prove that ACP = QCD.

Solution: ;
In triangles ACD and QCP,

A = P and Q = D [Angles in same segment]

ACD = QCP [Third angles] ……….(i)

Subtracting PCD from both the sides of eq. (i), we get,

ACD – PCD = QCP – PCD

ACPO = QCD

Hence proved.

Question 10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Solution:
Given: Two circles intersect each other at points A and B. AP and AQ be their respective diameters.

To prove: Point B lies on the third side PQ.

Construction : Join A and B.

Proof: AP is a diameter.

1 =

[Angle in semicircle]

AlsoAQ is a diameter.

2 =

[Angle in semicircle]

1 + 2 =

PBQ =

PBQ is a line.

Thus point B. i.e. point of intersection of these circles lies on the third side i.e., on PQ.

Question 11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that CAD = ABD.

Solution:
We have ABC and ADC two right triangles, right angled at B and D respectively.

ABC = ADC [Each ]

If we draw a circle with AC (the common hypotenuse) as diameter, this circle will definitely passes through of an arc AC, Because B and D are the points in the alternate segment of an arc AC.

Now we have subtending CBD and CAD in the same segment.

CAD = CBD

Hence proved.

 

NCERT Solutions for Class 9 Maths Exercise 10.6

Exercise 10.6 covers more complex problems on intersecting circles, parallel chords, and angle properties. It tests your ability to apply multiple circle theorems together. With NCERT Solutions for Class 9 Maths Exercise 10.6, students can confidently attempt class 9 maths chapter 10 exercises.

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’ respectively such that they intersect each other at P and Q.
To Prove: ∠OPO’ = ∠OQO’.
Construction : Join OP, O’P, OQ, O’Q and OO’.
Proof: In ∆OPO’ and ∆OQO’, we have

OP = OQ [Radii of the same circle]
O’P = O’Q [Radii of the same circle]
OO’ = OO’ [Common]
∴ AOPO’ = AOQO’ [By SSS congruence criteria]
⇒ ∠OPO’ = ∠OQO’ [C.P.C.T.]

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:
We have a circle with centre O.
AB || CD and the perpendicular distance between AB and CD is 6 cm and AB = 5 cm, CD = 11 cm.

Let r cm be the radius of the circle.
Let us draw OP ⊥ AB and OQ ⊥ CD such that
PQ = 6 cm
Join OA and OC.
Let OQ = x cm
∴ OP = (6 – x) cm
∵ The perpendicular drawn from the centre of a circle to chord bisects the chord.

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:
We have a circle with centre O. Parallel chords AB and CD are such that the smaller chord is 4 cm away from the centre.

Let r cm be the radius of the circle and draw OP ⊥ AB and join OA and OC.
∵ OP ⊥ AB
∴ P is the mid-point of AB.
⇒ AP = = = 3 cm
Similarly, CQ = = = 4 cm
Now in ∆OPA, we have OA2 = OP2 + AP2
⇒ r2 = 42 + 32
⇒ r2 = 16 + 9 = 25
⇒ r = =5
Again, in ∆CQO, we have OC2 = OQ2 + CQ2
⇒ r2 = OQ2 + 42
⇒ OQ2 = r2 – 42
⇒ OQ2 = 52 – 42 = 25 – 16 = 9 [∵ r = 5]
⇒ OQ
⇒ √9 = 3
The distance of the other chord (CD) from the centre is 3 cm.
Note: In case if we take the two parallel chords on either side of the centre, then

In ∆POA, OA ² = OP ² + PA ²
⇒ r ² = 42 + 3 ² = 5 ²
⇒ r = 5
In ∆QOC, OC ² = CQ ² + OQ ²
⇒ OQ ² = 4 ² + OQ ²
⇒ OQ ² = 5 ² – 4 ² = 9
⇒ OQ = 3

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:
Given : ∠ABC is such that when we produce arms BA and BC, they make two equal chords AD and CE.
To prove: ∠ABC = [∠DOE – ∠AOC]
Construction : Join AE.
Proof: An exterior angle of a triangle is equal to the sum of interior opposite angles.
∴ In ∆BAE, we have
∠DAE = ∠ABC + ∠AEC ……(i)
The chord DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle.

⇒ ∠ABC = [(Angle subtended by the chord DE at the centre) – (Angle subtended by the chord AC at the centre)]
⇒ ∠ABC = [Difference of the angles subtended by the chords DE and AC 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:
We have a rhombus ABCD such that its diagonals AC and BD intersect at O.
Taking AB as diameter, a circle is drawn. Let us draw PQ || DA and RS || AB, both are passing through O.
P, Q, R and S are the mid-points of DC, AB, AD and BC respectively,
∵ Q is the mid-point of AB.
⇒ AQ = QB …(i)
Since AD = BC [ ∵ ABCD is a rhombus]
∴ AD = BC
⇒ RA = SB
⇒ RA = OQ …(ii)

[ ∵ PQ is drawn parallel to AD and AD = BC]
We have, AB = AD [Sides of rhombus are equal]
⇒ AB = AD
⇒ AQ = AR …(iii)
From (i), (ii) and (iii), we have AQ = QB = OQ
i.e. A circle drawn with Q as centre, will pass through A, B and O.
Thus, the circle passes through the of intersection ‘O’ of the diagonals rhombus ABCD.

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:
We have a circle passing through A, B and C is drawn such that it intersects CD at E.
ABCE is a cyclic quadrilateral.
∴∠AEC + ∠B = 180° …(i)
[Opposite angles of a cyclic quadrilateral are supplementary] But ABCD is a parallelogram. [Given]
∴∠D = ∠B …(ii)
[Opposite angles of a parallelogram are equal]
From (i) and (ii), we have
∠AEC + ∠D = 180° …(iii)
But ∠AEC + ∠AED = 180° [Linear pair] …(iv)

From (iii) and (iv), we have ∠D = ∠AED
i.e., The base angles of AADE are equal.
∴ Opposite sides must be equal.
⇒ AE = AD

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:
Given: AC and BD of a circle bisect each other at O.

Then OA = OC and OB = OD

To prove: (i) AC and BD are the diameters. In other words, O is the centre of the circle.

(ii) ABCD is a rectangle.

Proof: (i) In triangles AOD and BOC,

AO = OC [given]

AOD = BOC [Vertically opp.]

OD = OB [given]

AOD COB [SAS congruency]

AD = CB [By CPCT]

Similarly AOB COD

AB = CD

[Arcs opposite to equal chords]

AC = BD [Chords opposites to equal arcs]

AC and BD are the diameters as only diameters can bisects each other as the chords of the circle.

(ii) Ac is the diameter. [Proved in (i)]

B = D = ….(i) [Angle in semi-circle]

Similarly BD is the diameter.

A = C = …(ii) [Angle in semi-circle]

Now diameters AC = BD

[Arcs opposite to equal chords]

AD = BC [Chords corresponding to the equal arcs] ….(iii)

Similarly AB = DC ….(iv)

From eq. (i), (ii), (iii) and (iv), we observe that each angle of the quadrilateral is and opposite sides are equal.

Hence ABCD is a rectangle.

Question 8. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that angles of the triangle are and respectively.

Solution:
According to question, AD is bisector of A.

1 = 2 =

And BE is the bisector of B.

3 = 4 =

Also CF is the bisector of C.

5 = 6 =

Since the angles in the same segment of a circle are equal.

9 = 3 [angles subtended by ] ….(i)

And 8 = 5 [angles subtended by ] ….(ii)

Adding both equations,

9 + 8 = 3 + 5

D =

Similarly E =

And F =

In triangle DEF,

D + E + F =

D = ( E + F )

D =

D =

D = [ A + B + C = ]

D =

Similarly, we can prove that

E = and F =

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:
Given: Two equal circles intersect in A and B.

A straight line through A meets the circles in P and Q.

To prove: BP = BQ

Construction: Join A and B.

Proof: AB is a common chord and the circles are equal.

Arc about the common chord are equal, i.e.,

Since equal arcs of two equal circles subtend equal angles at any point on the remaining part of the circle, then we have,

1 = 2

In triangle PBQ,

1 = 2 [proved]

Sides opposite to equal angles of a triangle are equal.

Then we have, BP = BQ

Question 10. In any triangle ABC, if the angle bisector of and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC.

Solution:
Given: ABC is a triangle and a circle passes through its vertices.

Angle bisector of A and the perpendicular bisector (say ) of its opposite side BC intersect each other at a point P.

To prove: Circumcircle of triangle ABC also passes through point P.

Proof: Since any point on the perpendicular bisector is equidistant from the end points of the corresponding side,

BP = PC ….(i)

Also we have 1 = 2 [ AP is the bisector of A (given)] ….(ii)

From eq. (i) and (ii) we observe that equal line segments are subtending equal angles in the same segment i.e., at point A of circumcircle of ABC. Therefore BP and PC acts as chords of circumcircle of ABC and the corresponding congruent arcs and acts as parts of circumcircle. Hence point P lies on the circumcircle. In other words, points A, B, P and C are concyclic (proved).

Additional Resource and Notes for class 9 Maths

1. Probability

2. Surface Areas and Volumes

3. Coordinate Geometry

4. Pair of Linear Equations in Two Variables

5. Line and angle

6 Number system

7. Mensuration

8. consutration

9. circle

10. Areas of parallelograms

11. Polynomial

12. Quadrilaterial

13. Triangle

14. Statistics

15. euclid geometr