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