Question

Prove that the bisectors of the four interior angles of a quadrilateral form a cyclic quadrilateral.

Answer 1

In $△DRC,\angle DRC+\angle RCD+\angle CDR={180}^{\circ }$ ......(i)

In$△APB,\angle APB+\angle PBA+\angle BAP={180}^{\circ }$ ......(ii)

Adding eq,(i) and (ii)we get

$\angle DRC+\angle RCD+\angle CDR+\angle APB+\angle PBA+\angle BAP={360}^{\circ }$......(iii)

$\Rightarrow \angle DRC+\frac{1}{2}\angle C+\frac{1}{2}\angle D+\angle APB+\frac{1}{2}\angle B+\frac{1}{2}\angle A={360}^{\circ }$

$\Rightarrow (\angle DRC+\angle APB)+\frac{1}{2}\times {360}^{\circ }={360}^{\circ }$

But $\angle DRC+\angle QRS$ ....(Vertically opposite angles)

and $\angle APB=\angle QPS$

$\therefore \angle QRS+\angle QPS={180}^{\circ }$

Thus, $PQRS$ is a cyclic quadrilateral.