Question

Question 7
If bisectors of $A\angle$ 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

Given, ABCD is a quadrilateral and all angles bisectors form a quadrilateral PQRS.

We know that, sum of all angles in a quadrilateral is ${360}^{\circ }$.
$\therefore \angle A+\angle B+\angle C+\angle D={360}^{\circ }.$
On dividing both sides by 2 , we get
$\frac{1}{2}(\angle A+\angle B+\angle C+\angle D)=\frac{{360}^{\circ }}{2}\Rightarrow \angle PAB+\angle PBA+\angle RCD+\angle RDC={180}^{\circ }.$
[Since, AP and PB are the bisectors of $\angle Aand\angle B$ respectively also RC and RD are the bisectors of $\angle C$ and $\angle D$ respectively]
Now, in $\Delta APB.$
$\angle PAB+\angle ABP+\angle BPA={180}^{\circ }$ ...(i)
[By angle sum property of a triangle]
$\Rightarrow \angle RDC+\angle DCR={180}^{\circ }-\angle CRD$
On substituting the value eqs. (ii) and (iii) in Eq. (i) we get
${180}^{\circ }-\angle BPA+{180}^{\circ }-\angle DRC={180}^{\circ }$
$\Rightarrow \angle BPA+{180}^{\circ }-\angle DRC={180}^{\circ }$
$\Rightarrow \angle SPQ+\angle SRQ={180}^{\circ }[\because \angle BPA=\angle SPQandDRC=\angle SRQ\text{vertically opposite angles}]$
Hence, PQRS is a quadrilateral whose opposite angle are supplementary.