Question

State true or false:

The bisectors of interior angles of a parallelogram doesn't form a rectangle.

• A. True
• B. False

Answer 1

The correct option is B False

Let $P,Q,R$ & $S$ be the points of intersection of the bisectors of $\angle A,\angle B,\angle C\&\angle D$ respectively of parallelogram ABCD.

Since DS bisects $\angle D$ & AS bisects $\angle A$ so $\angle DAB=2\angle DAS\&\angle ADC=2\angle ADS$

Here $\angle DAB+\angle ADC={180}^{\circ }$ [ adjacent angles of the parallelogram$={180}^{\circ }$]

$\Rightarrow \angle DAB+\angle ADC={180}^{\circ }$

$\Rightarrow 2\angle DAS+2\angle ADS={180}^{\circ }$

$\Rightarrow \angle DAS+\angle ADS={90}^{\circ }$ ---$\left(1\right)$

In $△ASD$,

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

$\Rightarrow \angle ASD+{90}^{\circ }={180}^{\circ }$ [ from $\left(1\right)$ ] $\Rightarrow \angle ASD={90}^{\circ }$

$\Rightarrow \angle PSR={90}^{\circ }$ [ $\angle ASD\&\angle PSR$ are vertically opposite angles ]

Similarly it can be shown that $\angle APB=\angle SPQ={90}^{\circ },\angle PQR={90}^{\circ }\&\angle SRQ={90}^{\circ }$.

So in quadrilateral PQRS all interior angles are right angles.

Hence $PQRS$ is a rectangle.