Question

In a parallelogram, the bisector of any two consecutive angles intersect at right angle, Prove it.

Answer 1

Let ABCD is a parallelogram

as we know

$\begin{array}{c}\angle A+\angle B={180}^0\\ OAbi\sec ts\angle DAB\&OBbi\sec ts\angle CBA\\ toprove\angle AOB={90}^0\\ nowin\Delta AOB\\ \angle OAB+\angle CBA+\angle AOB={180}^0\\ \frac{1}{2}\angle DAB+\frac{1}{2}\angle CBA+\angle AOB={180}^0\\ \frac{1}{2}\left(\angle DAB+CBA\right)+\angle AOB={180}^0\\ \frac{1}{2}\times {180}^0+\angle AOB={180}^0\\ {90}^0+\angle AOB={180}^0\\ \angle AOB={180}^0-{90}^0\\ \angle AOB={90}^0\end{array}$