Question

If bisectors of $\angle A$ and $\angle B$ of a quadrilateral $ABCD$ intersect each other at $P,$ of $\angle B$ and $\angle C$ at $Q.$ $\angle C$ and $\angle D$ at $R,$ $\angle D$ and $\angle A$ at $S$ then show that $PQRS$ is a quadrilateral whose opposite angle are supplementary

Answer 1

To Prove:

$\angle PSR+\angle PQR={180}^{\circ }\angle SPQ+\angle SRQ={180}^{\circ }$

Apply angle sum property in $△DSA,$

$\angle DAS+\angle ADS+\angle DSA={180}^{\circ }$

$\Rightarrow \frac{1}{2}\angle A+\frac{1}{2}\angle D+\angle DSA={180}^{\circ }$

$\Rightarrow \angle DSA={180}^{\circ }-\frac{1}{2}\angle A-\frac{1}{2}\angle D$

As, $\angle DSA$ and $\angle PSR$ are vertically opposite angle,

$\angle PSR={180}^{\circ }-\frac{1}{2}\angle A-\frac{1}{2}\angle D\dots \left(i\right)$

Similarly, in $△CQB,$

$\angle PQR={180}^{\circ }-\frac{1}{2}\angle C-\frac{1}{2}\angle B\dots \left(ii\right)$

On adding equations $\left(i\right)$ and $\left(ii\right),$ we get,

$\begin{array}{cc}\angle PSR+\angle PQR& ={180}^{\circ }-\frac{1}{2}\angle C-\frac{1}{2}\angle B+{180}^{\circ }-\frac{1}{2}\angle A-\frac{1}{2}\angle D\\ & ={360}^{\circ }-\frac{1}{2}\times \left(\angle A+\angle B+\angle C+\angle D\right)\\ & ={360}^{\circ }-\frac{1}{2}\times {360}^{\circ }\\ & ={180}^{\circ }\end{array}$

$\therefore \angle PSR+\angle PQR={180}^{\circ }$

In quadrilateral $PQRS,$

$\angle SPQ+\angle SRQ+\angle PSR+\angle PQR={360}^{\circ }$

$\Rightarrow \angle SPQ+\angle SRQ+{180}^{\circ }={360}^{\circ }$

$\Rightarrow \angle SPQ+\angle SRQ={180}^{\circ }$

Hence, proved that opposite angles of $PQRS$ are supplementary.