Question

In a parallelogram $ABCD$, the bisectors of the consecutive angles $\angle A$ and $\angle B$ intersect at $P$. Show that $\angle APB={90}^{\circ }$

Answer 1

Given parallelogram ABCD ,the bisectors of consecutive angles $\angle A$ and $\angle B$ intersect at P

The $AD\parallel BC$ and AB traversal them

$\therefore \angle DAB+\angle CBA={180}^0$

$\Rightarrow \frac{1}{2}\angle DAB+\frac{1}{2}\angle CBA={90}^0$

,the bisectors of consecutive angles $\angle A$ and $\angle B$ intersect at P

Then $\angle PAB=\frac{1}{2}\angle DAB$ and $\angle PBA=\frac{1}{2}\angle CBA$

$\therefore \angle PAB+\angle PBA={90}^0$

In $\Delta ABP$

$\angle APB+\angle PBA+\angle PAB={180}^0$

$\Rightarrow \angle APB+{90}^0={180}^0$

$\Rightarrow \angle APB=180-90={90}^0$