91Ó°ÊÓ

The figure shows quad. EFGH is a parallelogram.

If two parallel lines are intersected by a transversal then the angle made by a transversal with respect to two lines are called alternate interior angles which are equal in measures.

From the figure, it can be observed that, $\stackrel{¯}{EF}âˆ\yen \stackrel{¯}{HG},\stackrel{¯}{EG}$ is a transversal and $∠1,∠2,∠3,∠4$ are alternate interior angles such that

$\begin{array}{l}∠1â‰\dots ∠2;\\ ∠3â‰\dots ∠4\end{array}$

Consider $\mathrm{Δ}GHE$and $\mathrm{Δ}GFE$, where $∠1â‰\dots ∠2$, $∠3â‰\dots ∠4$and $\stackrel{¯}{EG}$ is common side in both the triangles therefore by ASA postulate, $\mathrm{Δ}GHEâ‰\dots \mathrm{Δ}GFE$.

As $\mathrm{Δ}GHEâ‰\dots \mathrm{Δ}GFE$, the sides of congruent triangles are congruent, that is, $\stackrel{¯}{HE}â‰\dots \stackrel{¯}{GF}$and $\stackrel{¯}{HG}â‰\dots \stackrel{¯}{EF}$.

Also $\stackrel{¯}{HE},\stackrel{¯}{GF}$and $\stackrel{¯}{HG},\stackrel{¯}{EF}$are opposite sides of a parallelogram.

Therefore, it is proved that the opposite sides of a parallelogram are congruent.