91Ó°ÊÓ

Consider the figure.

Consider the quadrilateral$PROV$.

Since the opposite sides of a quadrilateral are parallel then it is a parallelogram.

Here consider that$\stackrel{¯}{PR}\left|\right|\stackrel{¯}{VO};\stackrel{¯}{RO}\left|\right|\stackrel{¯}{PV}$ thus$PROV$ is a parallelogram.

Also$\stackrel{¯}{PR}â‰\dots \stackrel{¯}{RO}$ thus the parallelogram$PROV$ is a rhombus.

From the figure$\stackrel{¯}{RV}âŠ\yen \stackrel{¯}{PO}$, so that $∠REO={90}^{∘}$.

Consider the triangle$∠REO$.

Since the sum of all the angles in a triangle is 180 degrees thus

$\begin{array}{l}m∠REO+m∠EOR+m∠ORE={180}^{∘}\\ 90+m∠EOR+m∠2=180\\ m∠EOR+m∠2=180−90\\ m∠EOR+m∠2=90\text{(1)}\end{array}$

Also, from the figure$\stackrel{¯}{PR}\left|\right|\stackrel{¯}{VO};\stackrel{¯}{RO}\left|\right|\stackrel{¯}{PV}$and $\stackrel{¯}{RV}$is a transversal therefore

$\begin{array}{l}m∠PRVâ‰\dots m∠RVO\\ m∠PRV=m∠1\text{(2)}\end{array}$

The diagonal$\stackrel{¯}{RV}$bisects the angle$∠PRO$.

That is$m∠PRV=m∠VRO\text{(3)}$

Use (1), (2) and (3) as follows:

$m∠1+m∠2=90$

Therefore, the angles are complementary.