S Chand Class 10 CHAPTER 14 Circle Exercise 14 B

 Exercise 14 B

Question 1

Ans: In the figure in $\triangle A B C, A B=A C \quad \times XY|| B C$

To prove : $B C Y X$ is a cyclic quadrilateral proof: In $\triangle A B C$,

$\begin{aligned}&\text { XY } || B C \\&\text { So } \angle A X Y=\angle A B C \\&\text { But } \angle A B C=\angle A C B \\&\text { So } \angle A XY=\angle A C B\end{aligned}$

But Ext. $\angle A x y$ is equal to its interior opp. $\angle A C B$ So $B C YX$ is a cyclic quadrilateral.

Question 2

Ans: In the figure in quad. ABCD BC is produced to X Ext. $\angle X C D=$ int OPP. $\angle A$

To prove: Quad. $A B C D$ is a cyclic

Proof: $\angle X C D+\angle D C B=180^{\circ}$

$\angle A=\angle D C B=180^{\circ}$

So Quad $A B C D$ is a cyclic. Hence proved

Question 3

Ans: In $\triangle P Q R, P Q=P R$

A circle passing through $Q$ and $R$, intersects $P Q$ and $P R$ at $S$ and T respectively. S is Joined

To prove: ST||QR 

Proof : In $\triangle P Q R$

$P Q=P R$

So $\angle Q=\angle R$........(i)

If SQRT is a cyclic quadrilateral 

So Ext $\angle S=$ int $\cdot O P P \cdot \angle R$......(ii)

from ( i) and (ii)

$\angle Q=\angle S$

But these are corresponding angle 

So QR||ST 

Question 4

Ans: Two circles with center $\mathrm{O}_{1}$ and $\mathrm{O}_{2}$ Intersect each other at A and B 

$A O_{1} C$ and $A O_{2} D$ are diameters 

To prove: D, B and C are in a straight line or D, B and C are Collinear

Proof: If AC is the diameter of circle of center $\mathrm{O}_{1}$ 

So $\angle A B C=90^{\circ}$

similarly

$\angle A B D=90^{\circ}$

Adding we get 

$\angle A B C+\angle A B D=90^{\circ}+90^{\circ}=180^{\circ}$

So CBD is a straight line

Hence $D, B$ and $C$ are in the same straight line.

Hence proved 

Question 5

Ans: In circle with center O, AB is its diameter AD $\perp XY$ and $B C+ XY$ Which intersect the circle at E 

To prove: CE =AD 

Construction : Join A, E.

(IMAGE TO BE ADDED)

Proof: $\angle A E B=90^{\circ}$

So $\angle A E C=90^{\circ}$

But $\angle C=\angle D=90^{\circ}$

so $A E C D$ is a rectangle

if Opposite side of a rectangle are equal

So $\quad \angle E=A D$

Question 6

Ans: In a circle with center O, AC is its Diameter and Chord AB||CD

To prove: AB= CD 

Construction : 

(IMAGE TO BE ADDED)

Proof: In $\triangle A O B$ and $\triangle C O D$.

$\begin{aligned}&O A=O C \\&O B=O D \\&\angle B A O=\angle O C D \\&\text { SO } \triangle A O B \cong \triangle C O D \\&\text { SO } A B=C D\end{aligned}$

Question 7

Ans: (IMAGE TO BE ADDED)

Two circles circles intersect each other at $P$ and $Q$ Lines $A P B$ and $C Q D$ are drawn from the point of intersection Respectively P Q, A C, BD and, $A Q, O B$ CP and $P D$ are Joined.

To prove: 

(i) AC||BD (ii) $\angle C P D=\angle A Q B$

Proof: (i) If APQC is a cyclic quad 

So Ext. $\angle B P Q=$ int OPP $\angle C$ ..............(i)

If PBDQ is a cyclic quad.

So $\quad \angle B P Q+\angle D=180^{\circ}$

$\Rightarrow \angle C+\angle D=180^{\circ}$

But These are Co-interior angles 

So AC||BQ

(ii) In $\triangle A Q B$ and $\triangle C P D$,

$\angle P A Q=P C Q$

$\angle P B Q=\angle P D Q$

$\angle A B Q=\angle P D C$

So $\triangle A Q B \sim \triangle A P D$

So Third angler $=$ Third angle

$\Rightarrow \angle A Q B=\angle C P D$

Hence proved 

Question 8

Ans: (a) ABCD is a rhombus whose diagonals AC and BD intersects each other at O. A circle With AB as diameter is drawn 

(IMAGE TO BE ADDED)

To prove: The circle passes through O 

Proof: Let the circle drawn on AB as diameter does not passes through O, Let it intersect AC at P 

Join PB 

The  $\angle A P B=90^{\circ}$

 But $\angle A O B=90^{\circ}$

So $\angle A P B=\angle A O B$

But it is not possible because  $\angle A P B$ is the exterior angle of Triangle OPB and an Exterior angle of a triangle is always greater than its interior opposite angle So our supposition is wrong 

Hence the circle will pass through O 

Question 9

Ans: $A n$ isosceles trapezium $A B C D$ in which $A D \| B C$ and

A B=D C

To prove: $A B C D$ is cyclic

Construction: Draw AE and DF perpendicular on BC 

Proof: In right $\triangle A B E$ and $\triangle D C F$

Hyp. $A B=D C$ 

side $A E=D F$

$\text { So } \triangle A B E \cong \triangle D C F$

So $\angle B=\angle C$

Now AD $\| B C$

So $\angle D A B+\angle B=180^{\circ}$

$\Rightarrow \angle D A B+\angle C=180^{\circ}$

But here are sum of opposite angles of a quad 

So ABCD is a circle 

Question 10

Ans: PQRS is a cyclic quadrilateral

$\angle A, \angle B, \angle C$ and $\angle D$ are angles in the four segment so formed exterior to the cyclic quadrilateral

To prove: $\angle A+\angle B+\angle C+\angle D=6$ right angles

constructions: Join $A S$ and $A R$

Proof: In cyclic quad .$AS D P$

$\angle P A S+\angle D=2 \mathrm{rt}$ angles...........(i)

Similarly in cyclic quad ARBQ

$\angle R A Q+\angle B=2 \mathrm{rt}$. angles.......(ii)

and In cyclic quad .ARCS 

$\angle S A R+\angle C=2 rt$. angles.........(iii)

Adding (i), (ii)and (iii)

$\angle P A S+\angle D+\angle R A Q+\angle B+\angle S A R+\angle C$

$=2+2+2=6 \mathrm{rt}$. angles

$\Rightarrow \angle P A S+\angle S A R+\angle R A Q+\angle B+\angle C+\angle D$

=6 rt-angles

$\Rightarrow \angle A+\angle B+\angle C+\angle D=6 \mathrm{rt.}$ angles

Question 11

Ans: O is the circumcenter of $\triangle A B C O D \perp B C \cdot O B$ and $O C$ are Joined

To prove: $\angle B O D=\angle A$

Proof: arc $B C$ subtends $\angle B O C$ at the center

 and $\angle B A C$ at the remaining part of the circle

So $\angle B O C=2 \angle B A C$.......(i)

In $\triangle O B C, O D \perp B C$

O B=O C

So OD bisects $\angle B O C$

$\Rightarrow \angle B O D=\frac{1}{2} \angle B O C$...........(ii)

From in and (ii)

$\angle B O D=\frac{1}{2} \times 2 \angle B A C=\angle B A C$

Hence proved.

Question 12

Ans: ABCD is a cyclic quadrilateral 

A circle passing through A and B meet AD and BC at E and F respectively EF is joined 

To prove: EF||DC 

Proof : IF ABCD is a cyclic quad.

So $\angle 1+\angle 3=180^{\circ}$.....(i)

Similarly ABFE is a cyclic quadrilateral 

So $\angle 1+\angle 2=180^{\circ}$........(ii)

from (i) and (ii)

$\begin{aligned}& \angle 1+\angle 3=\angle L+\angle 2 \\\Rightarrow & \angle 3=\angle 2\end{aligned}$

But these are corresponding angles 

So EF || DC Hence proved