NCERT Exemplar for Class 9 Maths Chapter 10 - Circles (Book Solutions)

Multiple Choice Questions

Sample Question 1: In Fig. two congruent circles have centers $O$ and $O\text{'}$. Arc $AXB$ subtends an angle of $75^{\circ }$ at the center $O$ and arc $A^{\prime }YB\text{'}$ subtends an angle of $25^{\circ }$ at the centre $O\text{'}$. Then the ratio of arcs $AXB$ and $A^{\prime }YB\text{'}$ is: 

(A) $2:1$                                    

(B) $1:2$                            

(C) $3:1$                                

(D) $1:3$

Ans: Correct option is (C).

For the case 1:

Substituting the values of $\theta$

$\text{Area }(AXB)={\displaystyle \frac{1}{2}}{r}^2\left({75}^{\circ }\right)$

For the case 2:

Substituting the values of $\theta$

$\text{Area }\left(A^{\prime }Y{B}^{\prime }\right)={\displaystyle \frac{1}{2}}{r}^2\left({25}^{\circ }\right)$

Now, finding ratio

${\displaystyle \frac{\text{ Area }(AXB)}{\text{Area }\left(A^{\prime }Y{B}^{\prime }\right)}}$

$={\displaystyle \frac{{\displaystyle \frac{1}{2}}{r}^2(75)}{{\displaystyle \frac{1}{2}}{r}^2(25)}}={\displaystyle \frac{75}{25}}={\displaystyle \frac{3}{1}}$ 

So, the ratio is $3:1.$ 

Sample Question 2: In Fig. $AB$ and $CD$ are two equal chords of a circle with center O. $OP$ and $OQ$ are perpendiculars on chords $\text{AB}$ and $CD$, respectively. If $\mathrm{\angle }POQ=150^{\circ }$, then $\mathrm{\angle }APQ$ is equal to

(A) $30^{\circ }$                        

(B) $75^{\circ }$                     

(C) $15^{\circ }$                       

(D) $60^{\circ }$

Ans: Correct option is (B).

Assume $\mathrm{\angle }OPQ=\mathrm{\angle }1\text{and}\mathrm{\angle }OQP=\mathrm{\angle }2$

$\because AB=CD$

$OP=OQ$ (Equal chords are equidistant from the centre)

$\mathrm{\angle }1=\mathrm{\angle }2$ (Angles opposite to equal sides are equal) 

In POQ,

$\mathrm{\angle }1+\mathrm{\angle }2+\mathrm{\angle }POQ={180}^{\circ }$

$\mathrm{\angle }1+\mathrm{\angle }1+{150}^{\circ }={180}^{\circ }$

$2\mathrm{\angle }1={180}^{\circ }-{150}^{\circ }$

$2\mathrm{\angle }1={30}^{\circ }$

$\mathrm{\angle }1={15}^{\circ }$ 

$\text{APB}$ is a line segment

$\mathrm{\angle }BPO+\mathrm{\angle }1+\mathrm{\angle }APQ={180}^{\circ }$

${90}^{\circ }+{15}^{\circ }+\mathrm{\angle }APQ={180}^{\circ }$

$\mathrm{\angle }APQ={75}^{\circ }$ 

EXERCISE 10.1

1. $AD$ is the diameter of a circle and $AB$ is a chord. If $AD=34cm,AB=30cm$, the distance of $AB$ from the center of the circle is

(A) $17cm$  

(B) $15cm$                     

(C) $4cm$                       

(D) $8cm$

Ans: Correct option is (D).

Drawing $\text{OL}\mathrm{\perp }\text{AB}$.

$\because$ The perpendicular from the centre of a circle to a chord bisects the chord.

$\therefore AL=LB={\displaystyle \frac{1}{2}}AB=15cm$

Applying Pythagoras theorem in $\mathrm{\Delta }OLA$ 

${OA}^2={OL}^2+{AL}^2$

$(17{)}^2=O{L}^2+(15{)}^2$

$289=O{L}^2+225$

$O{L}^2=289-225=64$

$OL=8cm$ 

2. In figure, if $OA=5cm,AB=8cm$ and OD is perpendicular to AB then CD is equal to

(A) 2cm                   

(B) 3cm                    

(C) 4cm                            

(D) 5cm

Ans: Correct option is (A).

From the figure

$AC=CB={\displaystyle \frac{1}{2}}AB$

${\displaystyle \frac{1}{2}}\times 8=4cm$ 

Applying Pythagoras theorem in $\mathrm{\Delta }ACO$ 

$A{O}^2=A{C}^2+O{C}^2$

$(5{)}^2=(4{)}^2+O{C}^2$

$25=16+O{C}^2$

$O{C}^2=25-16=9$

$OC=3cm$ 

Hence,

$CD=OD-OC$

$=5-3=2cm$ 

3. If $AB=12cm,BC=16cm$ and $AB$ is perpendicular to $BC$, then the radius of the circle passing through the points $A,B$ and $C$ is

(A) $6cm$    

(B) $8cm$ 

(C) $10cm$                   

(D) $12cm$

Ans: Correct option is (C)

Here, $\text{BC}\mathrm{\perp }\text{AB}$, which implies it is a circle. 

Applying Pythagoras theorem for calculating diameter $\text{AC}$  

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=(12{)}^2+(16{)}^2$

$A{C}^2=144+256$

$A{C}^2=400$

$AC=\sqrt{400}=20cm$ 

Radius of circle$={\displaystyle \frac{1}{2}}(AC)$

${\displaystyle \frac{1}{2}}\times 20=10cm$

Hence, the radius of the circle is $10\text{cm}$.

4. In figure, if $\mathrm{\angle }ABC=20^{\circ }$, then $\mathrm{\angle }AOC$ is equal to 

(A) $20^{\circ }$                

(B) $40^{\circ }$                   

(C) $60^{{}^{\circ }}$                  

(D) $10^{\circ }$ 

Ans: Correct option is (B)

From the theorem that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

$\mathrm{\angle }AOC=2\mathrm{\angle }ABC$

$\mathrm{\angle }AOC=2\times 20^{\circ }=40^{\circ }$ 

5. In figure, if $AOB$ is a diameter of the circle and $AC=BC$, then $\mathrm{\angle }CAB$ is equal to

(A) $30^{\circ }$               

(B) $60^{\circ }$             

(C) $90^{\circ }$                

(D) $45^{\circ }$ 

Ans: Correct option is (D)

Here, $\mathrm{\angle }BCA=90^{\circ }$

$\mathrm{\angle }ABC=\mathrm{\angle }CAB$

$\because \text{[Angles opposite to equal sides are equal]}$

In $\mathrm{\Delta }ABC$,  

$\mathrm{\angle }CAB+\mathrm{\angle }ABC+\mathrm{\angle }BCA=180^{\circ }$(Sum of angles of triangle)

$\mathrm{\angle }CAB+\mathrm{\angle }CAB+\mathrm{\angle }90^{\circ }=180^{\circ }$

Now,

$2\mathrm{\angle }CAB=180^{\circ }-90^{\circ }$

$\mathrm{\angle }CAB={\displaystyle \frac{90^{\circ }}{2}}$

$\mathrm{\angle }CAB=45^{\circ }$ 

6. In figure, if $\mathrm{\angle }OAB=40^{\circ }$, and then $\mathrm{\angle }ACB$ is equal to 

(A) ${50}^{\circ }$            

(B) $40^{\circ }$                

(C) $60^{\circ }$                 

(D) ${70}^{\circ }$ 

Ans: Correct option is (A)

Here, In $\mathrm{\Delta }OAB$ 

$OA=OB$ (Radius of a circle)

$\because$ Angles opposite to equal sides are equal

$\mathrm{\angle }OAB=\mathrm{\angle }OBA$

$\mathrm{\angle }OBA=40^{\circ }$ 

Also, $\mathrm{\angle }AOB+\mathrm{\angle }OBA+\mathrm{\angle }BAO={180}^{\circ }$ (Sum of angles of triangle)

$\mathrm{\angle }AOB+40^{\circ }+40^{\circ }=180^{\circ }$

$\mathrm{\angle }AOB=180^{\circ }-80^{\circ }$

$\mathrm{\angle }AOB=100^{\circ }$ 

$\because$ Angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle. 

$\mathrm{\angle }AOB=2\mathrm{\angle }ACB$

$100^{\circ }=2\mathrm{\angle }ACB$

$\mathrm{\angle }ACB={\displaystyle \frac{100^{\circ }}{2}}=50^{\circ }$

$\mathrm{\angle }ACB=50^{\circ }$ 

7. In figure, if $\mathrm{\angle }DAB=60^{\circ },\mathrm{\angle }ABD=50^{\circ }$, then $\mathrm{\angle }ACB$ is equal to

(A) $60^{\circ }$                

(B) $50^{\circ }$                  

(C) $70^{\circ }$                    

(D) $80^{\circ }$ 

Ans: Correct option is (C) 

From the theorem,

$\mathrm{\angle }ADB=\mathrm{\angle }ACB$  (Angles in same segment of a circle are equal)

In $\mathrm{\Delta }ABD$ , using angle sum property of a triangle

$\mathrm{\angle }ABD+\mathrm{\angle }ADB+\mathrm{\angle }DAB=180^{\circ }$

$50^{\circ }+\mathrm{\angle }ADB+60^{\circ }=180^{\circ }$

$\mathrm{\angle }ADB=180^{\circ }-110^{\circ }$

$\mathrm{\angle }ADB=\mathrm{\angle }ACB=70^{\circ }$ 

8. $\text{ABCD}$ is a cyclic quadrilateral such that $AB$ is a diameter of the circle circumscribing it and $\mathrm{\angle }ADC=140^{\circ }$, then $\mathrm{\angle }BAC$ is equal to

(A) $80^{\circ }$                

(B) $50^{\circ }$                  

(C) $40^{\circ }$                    

(D) $30^{\circ }$ 

Ans: Correct option is (B)

Using the property sum of the opposite angles in a cyclic quadrilateral is${180}^{\circ }$. $\mathrm{\angle }ADC+\mathrm{\angle }ABC={180}^{\circ }$

$140^{\circ }+\mathrm{\angle }ABC=180^{\circ }$

$\mathrm{\angle }ABC=180^{\circ }-140^{\circ }$

$\mathrm{\angle }ABC=40^{\circ }$ 

$\because \mathrm{\angle }ACB$ is an angle in a semi-circle. 

$\therefore \mathrm{\angle }ACB=90^{\circ }$

Using angle sum property of a triangle $\mathrm{\Delta }ABC$

$\mathrm{\angle }BAC+\mathrm{\angle }ACB+\mathrm{\angle }ABC=180^{\circ }$

$\mathrm{\angle }BAC+90^{\circ }+40^{\circ }=180^{\circ }$

$\mathrm{\angle }BAC=180^{\circ }-130^{\circ }=50^{\circ }$

$\mathrm{\angle }BAC=50^{\circ }$ 

9. In figure, $BC$ is the diameter of the circle and $\mathrm{\angle }BAO=60^{\circ }$. Then, $\mathrm{\angle }ADC$ is equal to 

(A) $30^{\circ }$                

(B) $45^{\circ }$                  

(C) $60^{\circ }$                    

(D) $120^{\circ }$ 

Ans: Correct option is (C)

Using the property angles opposite to equal sides are equal

Here, in $\mathrm{\Delta }AOB$

$\mathrm{\angle }OBA=\mathrm{\angle }BAO$

$\mathrm{\angle }OBA=60^{\circ }$

$\because$ Angles in the same segment AC are equal $\mathrm{\angle }ABC=\mathrm{\angle }ADC$

$\therefore \mathrm{\angle }ADC=60^{\circ }$

10. In figure, if $\mathrm{\angle }AOB=90^{\circ }$ and $\mathrm{\angle }ABC=30^{\circ }$, then $\mathrm{\angle }CAO$ is equal to

(A) $30^{\circ }$                

(B) $45^{\circ }$                  

(C) $90^{\circ }$                    

(D) $60^{\circ }$ 

Ans: Correct option is (D)

In $\mathrm{\Delta }OAB,$ Using angle sum property of a triangle

$\mathrm{\angle }OAB+\mathrm{\angle }ABO+\mathrm{\angle }BOA=180^{\circ }$

$\mathrm{\angle }OAB+\mathrm{\angle }OAB+90^{\circ }=180^{\circ }$

$2\mathrm{\angle }OAB=180^{\circ }-90^{\circ }$

$\mathrm{\angle }OAB={\displaystyle \frac{90^{\circ }}{2}}=45^{\circ }$

In $\mathrm{\Delta }ACB,$ Using angle sum property of a triangle $\mathrm{\Delta }ACB,$

$\mathrm{\angle }ACB+\mathrm{\angle }CBA+\mathrm{\angle }CAB=180^{\circ }$

$\because \mathrm{\angle }ACB={\displaystyle \frac{\mathrm{\angle }A0B}{2}}={\displaystyle \frac{90^{\circ }}{2}}=45^{\circ }$

$45^{\circ }+30^{\circ }+\mathrm{\angle }CAB=180^{\circ }$

$\mathrm{\angle }CAB=180^{\circ }-75^{\circ }=105^{\circ }$ 

Also, 

$\mathrm{\angle }CAO+\mathrm{\angle }OAB=105^{\circ }$

$\mathrm{\angle }CAO+45^{\circ }=105^{\circ }$

$\mathrm{\angle }CAO=105^{\circ }-45^{\circ }=60^{\circ }$ 

Short Answer Questions with Reasoning

Write True or False and justify your answer.

Sample Question 1: The angles subtended by a chord at any two points of a circle are equal.

Ans: False

If two points lie in the same segment (major or minor) only, then the angles will be equal otherwise they are not equal.

Sample Question 2: Two chords of a circle of lengths $10\text{ cm and }8\text{ cm}$are at the distances $8.0\text{ cm}$ and $3.5\text{ cm}$, respectively from the centre.

Ans: False

Here, the larger chord is at a smaller distance from the center.

EXERCISE 10.2

Write True or False and Justify Your Answer in Each of the Following:

1. Two chords $AB$ and $CD$ of a circle are each at distances $4cm$ from the center. Then $AB=CD$.

Ans: The statement is True

The chords equidistant from the center of the circle are equal in length, so the statement two chords AB and CD of a circle are each at distances 4cm from the center. Then AB=CD.

2. Two chords $AB$ and $AC$ of a circle with center $O$ are on the opposite sides of $OA$. Then $\mathrm{\angle }OAB=\mathrm{\angle }OAC$.

Ans: The statement is False

Joining $\text{OB}$ and $\text{OC}$.

In $\mathrm{\Delta }OAB$ and $\mathrm{\Delta }OAC$,

$OA=OA$      (Common side)

$OB=OC$ (Radius of circle)

Here, either any angle or third side is not congruent to $\mathrm{\Delta }OAC$.

$\therefore$ $\mathrm{\angle }OAB\ne \mathrm{\angle }OAC$

3. Two congruent circles with centres $O$ and $O\text{'}$ intersect at two points $A$ and $B$. Then $\mathrm{\angle }AOB=\mathrm{\angle }A{O}^{\prime }B$.

Ans: The statement is True.

Joining $\text{AB,}\text{OA}$ and $\text{OB,}\text{O'A}$ and $\text{BO'}$.

In $\mathrm{\Delta }AOB$ and $\mathrm{\Delta }A{O}^{\prime }B$ 

$OA=A{O}^{\prime }$ (Same radius)

$OB=B{O}^{\prime }$ (Same radius) 

$AB=AB$ (Common chord)

$\mathrm{\Delta }AOB\cong \mathrm{\Delta }A{O}^{\prime }B$ (by SSS congruence rule)

$\mathrm{\angle }AOB=\mathrm{\angle }A{O}^{\prime }B$ (by CPCT)

4. Through three collinear points a circle can be drawn.

Ans: The statement is False

The circle can pass through only two collinear points but not through three collinear points. So, the statement is false.

5. A circle of radius $3cm$ can be drawn through two points $A,B$ such that $AB=6cm$.

Ans: The statement is True

diameter of a circle is $AB=6cm$ 

radius of a circle$={\displaystyle \frac{AB}{2}}={\displaystyle \frac{6}{2}}=3cm$, which is true.

6. If $AOB$ is the diameter of a circle and $\text{C}$ is a point on the circle, then $A{C}^2+B{C}^2=A{B}^2$.

Ans: The statement is True

Because any diameter of the circle subtends a right angle to any point on the circle. AOB is the diameter of a circle and C is a point on the circle. C is a right angle here.

Using Pythagoras theorem

In right angled $\mathrm{\Delta }ACB$

$A{C}^2+B{C}^2=A{B}^2$

7. $ABCD$ is a cyclic quadrilateral such that $\mathrm{\angle }A=90^{\circ },\mathrm{\angle }B=70^{\circ },\mathrm{\angle }C=95^{\circ }$ and $\mathrm{\angle }D=105^{\circ }$

Ans: The statement is False
For cyclic quadrilaterals, the sum of opposite angles is 180°.

Now calculating the sum of opposite angles of quadrilateral,

$\mathrm{\angle }A+\mathrm{\angle }C={90}^{\circ }+{95}^{\circ }={185}^{\circ }\ne {180}^{\circ }$
And $\mathrm{\angle }B+\mathrm{\angle }D={70}^{\circ }+{105}^{\circ }={175}^{\circ }\ne {180}^{\circ }$

So, it is not a cyclic quadrilateral.

8. If $A,B,C$ and $D$ are four points such that $\mathrm{\angle }BAC=30^{\circ }$ and $\mathrm{\angle }BDC=60^{\circ }$, then $D$ is the centre of the circle through $A,B$ and $C$.

Ans: The statement is False

Many points D can be there, such that $\mathrm{\angle }BDC=60^{\circ }$ and each such point cannot be the center of the circle through $A,B$ and $C$. So, the statement is false.

9. If  $A,B,C$ and $D$ are four points such that $\mathrm{\angle }BAC=45^{\circ }$ and $\mathrm{\angle }BDC=45^{\circ }$, then  $A,B,C$ and $D$ are concyclic.

Ans: The statement is True.
$\because \mathrm{\angle }BAC=45^{\circ }$ and $\mathrm{\angle }BDC=45^{\circ }$

Angles in the same segment of a circle are equal. Hence, n  $A,B,C$ and $D$ are Concyclic.

10.  In Fig. , if $AOB$ is a diameter and $\mathrm{\angle }ADC=120^{\circ }$, then $\mathrm{\angle }CAB=30^{\circ }$.

Ans: The statement is True

Since, ADCB is a cyclic quadrilateral.

$\mathrm{\angle }\mathrm{ADC}+\mathrm{\angle }\mathrm{CBA}={180}^{\circ }$.

(sum of opposite angles of cyclic quadrilateral is ${180}^{\circ }$ )

$\Rightarrow \mathrm{\angle }CBA={180}^{\circ }-{120}^{\circ }={60}^{\circ }[\because \mathrm{\angle }ADC={120}^{\circ }]$

In $\mathrm{△}ACB,\mathrm{\angle }CAB+\mathrm{\angle }CBA+\mathrm{\angle }ACB={180}^{\circ }$

(by angle sum property of a triangle)

$\mathrm{\angle }\mathrm{CAB}+{60}^{\circ }+{90}^{\circ }={180}^{\circ }$

(triangle formed from diameter to the circle is ${90}^{\circ }$ i.e., $\mathrm{\angle }ACB={90}^{\circ }$ )

$\Rightarrow \mathrm{\angle }CAB={180}^{\circ }-{150}^{\circ }={30}^{\circ }.$

Short Answer Questions

Sample Question 1: In Fig. AOC is the diameter of the circle and arc AXB = ${\displaystyle \frac{1}{2}}$ arc BYC. Find $\mathrm{\angle }BOC$

Ans: If two chords of a circle are equal, then their corresponding arcs are congruent and conversely, if two arcs are congruent, then their corresponding chords are equal.

As $\text{arc}AXB={\displaystyle \frac{1}{2}}\text{arc}BYC$

Also, $\mathrm{\angle }AOB+\mathrm{\angle }BOC={180}^{\circ }$

${\displaystyle \frac{1}{2}}\mathrm{\angle }BOC+\mathrm{\angle }BOC={180}^{\circ }$

$\mathrm{\angle }BOC={\displaystyle \frac{2}{3}}\times {180}^{\circ }={120}^{\circ }$ 

Sample Question 2: In Fig. $\mathrm{\angle }ABC=45^{\circ }$, prove that $OA\mathrm{\perp }OC$.

Ans: $\because \mathrm{\angle }ABC={\displaystyle \frac{1}{2}}\mathrm{\angle }AOC$

$\mathrm{\angle }AOC=2\mathrm{\angle }ABC=2\times {45}^{\circ }={90}^{\circ }$

$\mathrm{\angle }AOC=2\times {45}^{\circ }={90}^{\circ }$

$OA\mathrm{\perp }OC$ 

EXERCISE 10.3

1. If arcs $AXB$  and $CYD$  of a circle are congruent, find the ratio of $AB$ and $CD$. 

Ans: Assume $\text{AXB}$  and $\text{CYD}$ are arcs of circle     

Centre and radius are $\text{O}$ and $r$ units,

Hence, the ratio of $\text{AB}$ and $\text{CD}$ is $1:1$

2. If the perpendicular bisector of a chord  $AB$ of a circle $PXAQBY$  intersects the circle at $P$ and $Q$, prove that $arcPXA\cong arcPYB$. 

Ans: If two chords of a circle are equal, then their corresponding arcs are congruent and conversely, if two arcs are congruent, then their corresponding chords are equal.

Assume $AB$ be a chord of a circle with a center at $O$. PQ is the perpendicular bisector of the chord $AB$ which intersects at $\text{M}$ passing through $O$.

Joining  $AP\text{ and }BP$

In $\mathrm{\Delta }APM$ and$\mathrm{\Delta }BPM$, 

$AM=MB$

$\mathrm{\angle }PMA=\mathrm{\angle }PMB$

$PM=PM$ (Since, OM is the bisector of AB)

$\mathrm{\Delta }\text{APM}\cong \mathrm{\Delta }\text{BPM}$(SAS axiom of congruency)

$\therefore PA=PB$

$\therefore$ $\text{arc }PXA\cong \text{arc }PYB$

3. $A,B$, and $C$ are three points on a circle. Prove that the perpendicular bisectors of $AB,BC,andCA$ are concurrent. 

Ans: If$A,B$ and $C$ are three points on a circle then the perpendicular bisectors of $AB,BC\text{ and}CA$ are concurrent.

Drawing perpendicular bisectors of $AB$ meet at a point $O$. 

Joining $OA,OB\text{ and }OC$

$AE=BE$               ($E$ is the perpendicular bisector of $\text{AB}$ )

$\mathrm{\angle }AEO=\mathrm{\angle }BEO={90}^{\circ }$

$OE=OE(\text{common})$

$\mathrm{\Delta }OEB\cong \mathrm{\Delta }OEA$ (by SAS congruence rule)

$\therefore OA=OB$(By CPCT)

Similarly,

$\mathrm{\Delta }OFA\cong \mathrm{\Delta }OFB$

$\therefore OA=OC$

$\text{So, }OA=OB=OC=x\left(\text{say}\right)$

So, in $\mathrm{\Delta }OMBand\mathrm{\Delta }OMC$

$OB=OC$ (from above)

$OM=OM$ (common)

$\mathrm{\angle }OMB=\mathrm{\angle }OMC={90}^{\circ }$
$\therefore \mathrm{\Delta }OEB\cong \mathrm{\Delta }OEA$ (by RHS congruence rule)

$BM=MC$ (by CPCT)

Therefore, $OM$ is the perpendicular bisector of $BC$

Hence, $OL,ON\text{ and}OM$ are concurrent

Hence, proved.

4. $AB$ and $AC$ are two equal chords of a circle. Prove that the bisector of the angle $BAC$  passes through the center of the circle. 

Ans: Given: $AB$ and $CD$ are two equal chords of the circle.

Prove: Center O lies on the bisector of the $\mathrm{\angle }BAC$.

Construction: Join BC. Let the bisector of $\mathrm{\angle }BAC$ intersect $BC$ in $P$.

Proof:

In $\mathrm{△}\mathrm{APB}$ and $\mathrm{△}\mathrm{APC}$

$\mathrm{AB}=\mathrm{AC}\dots ($ Given $)$

$\mathrm{\angle }\mathrm{BAP}=\mathrm{\angle }\mathrm{CAP}\dots ($ Given $)$

$\mathrm{AP}=\mathrm{AP}\dots .($ common $)$

$\therefore \mathrm{△}\mathrm{APB}\cong \mathrm{△}\mathrm{APC}\dots \mathrm{SAS}$ test

$\Rightarrow \mathrm{BP}=\mathrm{CP}$ and $\mathrm{\angle }\mathrm{APB}=\mathrm{\angle }\mathrm{APC}\dots \mathrm{CPCT}$

$\mathrm{\angle }\mathrm{APB}+\mathrm{\angle }\mathrm{APC}={180}^{\circ }\dots ($ Linear pair $)$

$\therefore 2\mathrm{\angle }\mathrm{APB}={180}^{\circ }\dots .(\mathrm{\angle }\mathrm{APB}=\mathrm{\angle }\mathrm{APC})$

$\Rightarrow \mathrm{\angle }\mathrm{APB}={90}^{\circ }$

$\therefore$ AP is a perpendicular bisector of chord BC.

Hence, AP passes through the center of the circle.

5. If a line segment joining midpoints of two chords of a circle passes through the center of the circle, prove that the two chords are parallel.

Ans: If a line segment joining midpoints of two chords of a circle passes through the center of the circle then the two chords are parallel.

Assume $AB\text{ and}CD$ are two chords of a circle.

Here, the centre is $O$ and $PQ$ is a diameter bisecting the chord $AB\text{ and}CD$ at $L$ and $M$.

Diameter $PQ$ passes through the center $O$ of the circle

Proof:

$\because L$ is the midpoint of $AB$ 

$\therefore$$OL\mathrm{\perp }AB$   (the line joining the center of a circle to the midpoint of a chord is perpendicular to the chord)

$\mathrm{\angle }ALO={90}^{\circ }$

Similarly, $OM\mathrm{\perp }CD$

$\therefore$$\mathrm{\angle }OMD={90}^{\circ }$

Hence, $\mathrm{\angle }ALO=\mathrm{\angle }OMD={90}^{\circ }$

As alternating angles are equal that means chords are parallel.

6. $ABCD$ is such a quadrilateral that $A$ is the centre of the circle passing through $B,C$ and $D$. Prove that $\mathrm{\angle }CBD+\mathrm{\angle }CDB={\displaystyle \frac{1}{2}}\mathrm{\angle }BAD$

Ans:  Join $AC\text{ and }BD$

$\because \mathrm{arc}DC$ subtends $\mathrm{\angle }DAC$ at the centre and $\mathrm{\angle }CBD$ at a point $B$ in the remaining part of the circle.

$\therefore \mathrm{\angle }DAC=2\mathrm{\angle }CBD$$..................(i)$

Similarly, 

$\because$ Arc $BC$ subtends $\mathrm{\angle }CAB$ at the centre and $\mathrm{\angle }CDB$ at a point $D$ in the remaining part of the circle.

$\therefore \mathrm{\angle }CAB=2\mathrm{\angle }CDB.....................(ii)$

Adding $\text{equations }\left(i\right)\text{ and }(ii)$, we get

$\mathrm{\angle }DAC+\mathrm{\angle }CAB=2\mathrm{\angle }CBD+2\mathrm{\angle }CDB$

$\mathrm{\angle }BAD=2(\mathrm{\angle }CBD+\mathrm{\angle }CDB)$

$\mathrm{\angle }CDB+\mathrm{\angle }CBD={\displaystyle \frac{1}{2}}\mathrm{\angle }BAD$ 

Hence  proved.

7. $O$ is the circumcentre of the $\mathrm{\Delta }ABC$ and $D$ is the midpoint of the base $BC$. Prove that $\mathrm{\angle }BOD=\mathrm{\angle }A$. 

Ans: Joining  $OB,OD\text{ and }OC$

In $\mathrm{\Delta }BOD\text{ \& }\mathrm{\Delta }COD$

$OB=OC\left(\text{radius of same circle}\right)$

$BD=DC\left(\because OD\text{ is the bisector of BC}\right)$

$OD=OD\left(\text{common}\right)$

$\mathrm{\Delta }BOD\cong \mathrm{\Delta }COD$

$\therefore \mathrm{\angle }BOD=\mathrm{\angle }COD\text{( By}\text{CPCT)}$ 

Using the property the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

$2\mathrm{\angle }BAC=\mathrm{\angle }BOC$

$\mathrm{\angle }BAC={\displaystyle \frac{2}{2}}\mathrm{\angle }BOD$ 

$\because \mathrm{\angle }BOC=2\mathrm{\angle }BOD$

$\mathrm{\angle }BAC=\mathrm{\angle }BOD$ 

Hence proved.

8. On a common hypotenuse $AB$, two right-angled triangles $ACBandADB$ are situated on opposite sides. Prove that $\mathrm{\angle }BAC=\mathrm{\angle }BDC$.

Ans: Joining $CD$. 

Assume O be the midpoint of AB.

$OA=OB=OC=OD$

$\because$ Mid-point of the hypotenuse of a right triangle is equidistant from its vertices. 

Now, drawing a circle to pass through the points $A,B,C$ and $D$.with $O$ as centre and radius equal to $OA$.

From the figure,

$\mathrm{\angle }BAC$ and $\mathrm{\angle }BDC$                ( angles of same segment $BC.$)

$\therefore$ $\mathrm{\angle }BAC=\mathrm{\angle }BDC$ 

Hence proved.

9. Two chords $AB$ and $AC$ of a circle subtends angles equal to $90^{\circ }$ and $150^{\circ }$, respectively at the centre. Find $\mathrm{\angle }BAC$, if $AB$ and $AC$ lie on the opposite sides of the centre. 

Ans: In$\mathrm{\Delta }BOA$,

$OB=OA$ (Radius of circle)

$\therefore$$\mathrm{\angle }OAB=\mathrm{\angle }OBA$$\dots$(i)  (Angles opposite to equal sides are equal)

Using angle sum property of a triangle

$\mathrm{\angle }OAB+\mathrm{\angle }OAB+{90}^{\circ }={180}^{\circ }$

$2\mathrm{\angle }OAB={180}^{\circ }-{90}^{\circ }$

$\mathrm{\angle }OAB={\displaystyle \frac{{90}^{\circ }}{2}}={45}^{\circ }$ 

Now, in $\mathrm{\Delta }AOC$, 

$AO=OC$ (Radius of a circle)

$\mathrm{\angle }OCA=\mathrm{\angle }OAC\dots$(ii) (Angles opposite to equal sides are equal)

Using angle sum property of a triangle

$\mathrm{\angle }AOC+\mathrm{\angle }OAC+\mathrm{\angle }OCA={180}^{\circ }$

${150}^{\circ }+2\mathrm{\angle }OAC={180}^{\circ }.$

$2\mathrm{\angle }OAC={180}^{\circ }-{150}^{\circ }$

$2\mathrm{\angle }OAC={30}^{\circ }$

$\mathrm{\angle }OAC={15}^{\circ }$

$\text{Also}\mathrm{\angle }BAC=\mathrm{\angle }OAB+\mathrm{\angle }OAC$

$\mathrm{\angle }BAC={45}^{\circ }+{15}^{\circ }={60}^{\circ }$ 

10. If $BM$ and $CN$ are the perpendiculars drawn on the sides $AC$ and $AB$ of the $\mathrm{\Delta }ABC$, prove that the points $B,C$, $M$ and $N$ are concyclic. 

Ans: Drawing a circle passing through the points $B,C$, $M$ and $N$

$\because \mathrm{\angle }BNC=\mathrm{\angle }BMC={90}^{\circ }$ 

Angles subtended by the diameter in a semicircle $={90}^{\circ }$

So, BC is the diameter of the circle.

points $M$ and $N$ should be on the same circle.

Hence, BCMN form a cyclic quadrilateral.

$\therefore$ The points B,C,M and N are concyclic.

11. If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed, is cyclic.

Ans: Assume $\mathrm{\Delta }ABC$ is an isosceles triangle with$AB=AC$.

Drawing a circle passing through the points $B,C,D$ and$E$.

In $\mathrm{\Delta }ABC,$ 

$AB=AC$ (Equal sides of an isosceles triangle) 

$\mathrm{\angle }ACB=\mathrm{\angle }ABC$        ………….... (i) (Angles opposite to the equal sides are equal)

Since,  $\mathrm{\angle }ADE=\mathrm{\angle }ACB$       {Corresponding angles}          …………... (ii)

Adding both sides by $\mathrm{\angle }EDC$ in Eq. (ii), 

$\mathrm{\angle }ADE+\mathrm{\angle }EDC=\mathrm{\angle }ACB+\mathrm{\angle }EDC$

${180}^{\circ }=\mathrm{\angle }ACB+\mathrm{\angle }EDC$ 

Here,  $\mathrm{\angle }ADE$ and $\mathrm{\angle }EDC$ form linear pair 

$\mathrm{\angle }EDC+\mathrm{\angle }ABC={180}^{\circ }$ {From Eq.(i)}

Hence, $\text{BCDE}$ is a cyclic quadrilateral, as the sum of the opposite angles is ${180}^{\circ }$.

12. If a pair of opposite sides of a cyclic quadrilateral are equal, and then prove that its diagonals are also equal. 

Ans: Assume$\text{ABCD}$ be a cyclic quadrilateral and $AD=BC$.

Joining $AC$ and $BD$ 

In $\mathrm{\Delta }AOD$ and $\mathrm{\Delta }BOC$, 

$\mathrm{\angle }OAD=\mathrm{\angle }OBC$ and $\mathrm{\angle }ODA=\mathrm{\angle }OCB$(same segments subtends equal angle to the circle)

$\text{AD}=\text{BC}$ (Given)

$\mathrm{\Delta }AOD\cong \mathrm{\Delta }BOC$ (By ASA congruence rule)

Adding $\text{DOC}$ on both sides

$\mathrm{\Delta }AOD+\mathrm{\Delta }DOC\cong \mathrm{\Delta }BOC+\mathrm{\Delta }DOC$

$\mathrm{\Delta }ADC\cong \mathrm{\Delta }BCD$

$AC=BD\left[\text{by CPCT}\right]$ 

13. The circumcenter of the $\mathrm{\Delta }ABC$ is O. Prove that $\mathrm{\angle }OBC+\mathrm{\angle }BAC={90}^{\circ }.$

Ans: Joining $BO\text{ \& }CO$ .

Assume $\mathrm{\angle }OBC=\mathrm{\angle }OCB=\theta$

In $\mathrm{\Delta }OBC$ using angle sum property of a triangle 

$\mathrm{\angle }BOC+\mathrm{\angle }OCB+\mathrm{\angle }CBO={180}^{\circ }$

$\mathrm{\angle }BOC+\theta +\theta ={180}^{\circ }$

$\mathrm{\angle }BOC={180}^{\circ }-2\theta$

Also, $\mathrm{\angle }BOC=2\mathrm{\angle }BAC$

$\mathrm{\angle }BAC={\displaystyle \frac{\mathrm{\angle }BOC}{2}}$

$\mathrm{\angle }BAC={\displaystyle \frac{{180}^{\circ }-2\theta }{2}}$

$\mathrm{\angle }BAC={90}^{\circ }-\theta$

$\mathrm{\angle }BAC+\theta ={90}^{\circ }$

$\therefore \mathrm{\angle }BAC+\mathrm{\angle }OBC={90}^{\circ }$ 

Hence, proved.

14. A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in the major segment. 

Ans: Assume $AB$ is a chord of a circle,

As chord is equal to the radius of the circle

$AB=BO\dots$ (i) 

Joining $OA,AC\text{ and }BC$

$\because OA=OB=$ Radius of circle

$OA=AB=BO$

Thus, $\mathrm{\Delta }OAB$ is an equilateral triangle.

As Each angle of an equilateral triangle is ${60}^{\circ }$

$\mathrm{\angle }AOB={60}^{\circ }$ 

$\mathrm{\angle }AOB=2\mathrm{\angle }ACB$

$\mathrm{\angle }ACB={\displaystyle \frac{{60}^{\circ }}{2}}={30}^{\circ }$ 

15. In figure, $\mathrm{\angle }ADC=130^{\circ }$ and chord $BC=$ chord $BE$. Find $\mathrm{\angle }CBE$.

Ans: Using the property that the sum of opposite angles of a cyclic quadrilateral ${180}^{\circ }$. 

$\mathrm{\angle }ADC+\mathrm{\angle }OBC={180}^{\circ }$

${130}^{\circ }+\mathrm{\angle }OBC={180}^{\circ }$

$\mathrm{\angle }OBC={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

$\mathrm{\angle }OBC={50}^{\circ }$ 

In $\mathrm{\Delta }BOC$ and $\mathrm{\Delta }BOE$,

$BC=BE$ (Given equal chord) 

$OC=OE$ (Both are the radius of the circle) 

$OB=OB$ (Common side) 

$\mathrm{\Delta }BOC\cong \mathrm{\Delta }BOE$

Now, $\mathrm{\angle }OBC=\mathrm{\angle }OBE={50}^{\circ }$ (by CPCT)

$\mathrm{\angle }CBE=\mathrm{\angle }CBO+\mathrm{\angle }EBO$

$\mathrm{\angle }CBE={100}^{\circ }$ 

16. In Fig., $\mathrm{\angle }ACB=40^{\circ }$. Find $\mathrm{\angle }OAB$.

Ans: $\therefore \mathrm{\angle }AOB=2\mathrm{\angle }ACB$

$\mathrm{\angle }ACB={\displaystyle \frac{\mathrm{\angle }AOB}{2}}$

${40}^{\circ }={\displaystyle \frac{1}{2}}\mathrm{\angle }AOB$ 

$\mathrm{\angle }AOB={80}^{\circ }$     …………………….(i) 

In $\mathrm{\Delta }AOB,$                                                        

$AO=BO$                                                 (Both are the radius of a circle)

$\mathrm{\angle }OBA=\mathrm{\angle }OAB$   …………………….(ii)

Using angle sum property of triangle

$\therefore \mathrm{\angle }AOB+\mathrm{\angle }OBA+\mathrm{\angle }OAB={180}^{\circ }$ 

${80}^{\circ }+\mathrm{\angle }OAB+\mathrm{\angle }OAB={180}^{\circ }$               (From  Eqs. (i) and (ii))

$2\mathrm{\angle }OAB={180}^{\circ }-{80}^{\circ }$