Question

If bisector of $\angle A$ and $\angle B$ of a quadrilateral $ABCD$ intersect each other at $P$, of $\angle B$ and $\angle C$ at $Q$, of $\angle C$ and $\angle D$ at $R$ and of $\angle D$ and $\angle A$ at $S$, then $PQRS$ is a

• A. Rectangle
• B. Rhombus
• C. Parallelogram
• D. Quadrilateral whose opposite angles are supplementary

Answer 1

The correct option is D

Quadrilateral whose opposite angles are supplementary

Explanation for the correct option

We know that,

From angle sum property of quadrilateral some of the angle of a quadrilateral ${360}^{°}$

$\Rightarrow \angle A+\angle B+\angle C+\angle D=360°$

On dividing both side by $2$

$\Rightarrow \frac{1}{2}\left(\angle A+\angle B+\angle C+\angle D\right)=180°$

$\because AP,PB,RC$ and $RD$ are bisector of $\angle A$,$\angle B$,$\angle C$ and $\angle D$

$\Rightarrow \angle PAB+\angle ABP+\angle RCD+\angle RDC=180°...............\left(1\right)$

We also know that,

Sum of all angles of a triangle $=180°$

$$\begin{gathered} \Rightarrow \angle PAB+\angle APB+\angle ABP=180° \\ \Rightarrow \angle PAB+\angle ABP=180°-\angle APB..............\left(2\right) \end{gathered}$$

Also

$$\begin{gathered} \therefore \angle RDC+\angle RCD+\angle CRD=180° \\ \Rightarrow \angle RDC+\angle RCD=180°-\angle CRD..............\left(3\right) \end{gathered}$$

Substituting the value of equation $\left(2\right),\left(3\right)$ in equation $\left(1\right)$

$$\begin{gathered} 180°-\angle APB+180°-\angle CRD=180° \\ \Rightarrow 360°-\angle APB-\angle CRD=180° \\ \Rightarrow \angle APB+\angle CRD=360°-180° \\ \Rightarrow \angle APB+\angle CRD=180°...................\left(4\right) \end{gathered}$$

Now,

$\angle SPQ=\angle APB$

$\angle SRQ=\angle DRC$

Substituting in equation $\left(4\right)$

$\Rightarrow \angle SPQ+\angle SRQ=180°$

And,

$\angle PSR+\angle PQR=180°[\mathrm{Angle}\mathrm{sum}\mathrm{property}\mathrm{of}\mathrm{quadrilateral}]$

Therefore $PQRS$ is a quadrilateral whose opposite angles are supplementary.

Hence, option (D) is correct.