Prove Theorem $5-1$.

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, $\overline{EF}\parallel \overline{HG},\overline{EG}$ is a transversal and $\angle 1,\angle 2,\angle 3,\angle 4$ are alternate interior angles such that

$\begin{array}{l}\angle 1\cong \angle 2;\\ \angle 3\cong \angle 4\end{array}$

Consider $\Delta GHE$and $\Delta GFE$, where $\angle 1\cong \angle 2$, $\angle 3\cong \angle 4$and $\overline{EG}$ is common side in both the triangles therefore by ASA postulate, $\Delta GHE\cong \Delta GFE$.

As $\Delta GHE\cong \Delta GFE$, the sides of congruent triangles are congruent, that is, $\overline{HE}\cong \overline{GF}$and $\overline{HG}\cong \overline{EF}$.

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

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