Question

The bisectors of any two adjacent angles intersect at ${90}^{\circ }$

Answer 1

Let $ABCD$ be the parallelogram

$OA$ and $OB$ are the bisectors of $\angle A$ and $\angle B$

From the figure, $\angle A+\angle B={180}^o$ [co-interior angles]

$\frac{1}{2}\angle A+\frac{1}{2}\angle B={90}^o$

From $△AOB$

$\angle ABO+\angle BAO+\angle OAB={180}^o$ [angle sum property]

$\angle OAB={180}^o-(\angle BAO+\angle ABO)$

$\angle OAB={180}^o-{90}^o$ [above proved]

$\Rightarrow \angle OAB={90}^o$