Question

If bisectors of $\angle A$ and $\angle B$ of a quadrilateral ABCD intersect each other at $P$, bisectors of $\angle B$ and $\angle C$ at $Q$, bisectors of $\angle C$ and $\angle D$ at $R$ and bisectors 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

From question the diagram we get is above.

From figure,

$\angle QPS=\angle APB$ (Vertically opposite angles) ---- (1)

Consider, $△APB$

$\angle APB+\frac{1}{2}\angle A+\frac{1}{2}\angle B={180}^{\circ }$

$\angle APB={180}^{\circ }-\frac{1}{2}(\angle A+\angle B)$ ----- (2)

From equation (1) & (2),

$\angle QPS={180}^{\circ }-\frac{1}{2}(\angle A+\angle B)$ ----- (3)

Similarly, $\angle QRS={180}^{\circ }$ - $\frac{1}{2}(\angle C+\angle D)$ ---- (4)

From (3) &(4),

$\angle QPS+\angle QRS={360}^{\circ }$ - $\frac{1}{2}(\angle A+\angle B+\angle C+\angle D)$

$={360}^{\circ }-\frac{1}{2}\left({360}^{\circ }\right)$

$={360}^{\circ }-{180}^{\circ }$

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

Hence, PQRS is a quadrilateral whose opposite angles are supplementary.