Question

In the figure, $ABCD$ is a parallelogram in which the angle bisectors of $\angle A$ and $\angle B$ intersect at the point $P$. Prove that $\angle APB={90}^{\circ }$

Answer 1

Since, $ABCD$ is a Parallelogram.

Therefore,

$AD||BC$

$AB$ is a transversal.

Therefore,

$A+B=180°.\left[Consecutiveinteriorangles\right]$

Multiply both sides by $\frac{1}{2}$,

$\frac{1}{2}A+\frac{1}{2}B=\frac{1}{2}\times \left({180}^0\right)$

$\frac{1}{2}A+\frac{1}{2}B={90}^0$ $..........\left(1\right)$

Since, $AP$ and $PB$ are angle bisectors of $A$ and $B$.

Therefore,

$\angle 1=\frac{A}{2}$

$\angle 2=\frac{B}{2}$

Substitute the values in

$\angle 1+\angle 2={90}^0$ $.........\left(2\right)$

Now, in $△APB$,

$1+APB+2={180}^0$

${90}^0+APB={180}^0$ $Fromequation\left(2\right)$

$\angle APB={90}^0$

Hence, Proved.