Prove Theorem 5-2. (Draw and label a diagram. List what is given and what is to be proved.)

The figure shows quad. ABCD 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, $\overline{AB}\parallel \overline{CD}$, $\overline{BD}$is a transversal and $\angle ABD,\angle BDC$, $\angle ADB,\angle DBC$are alternate interior angles such that

$\begin{array}{l}\angle ABD\cong \angle BDC;\\ \angle ADB\cong \angle DBC\end{array}$

Consider $\Delta ADB$and $\Delta BCD$, where $\angle ABD\cong \angle BDC$, $\angle ADB\cong \angle DBC$and $\overline{BD}$is common side in both the triangles therefore by ASA postulate, $\Delta ADB\cong \Delta BCD$.

As $\Delta ADB\cong \Delta BCD$, the angles of congruent triangles are congruent, that is,$\angle A\cong \angle C$.

From the figure, it can be observed that, $\overline{AD}\parallel \overline{BC}$, $\overline{AC}$is a transversal and $\angle DCA,\angle CAB$, $\angle DAC,\angle ACB$are alternate interior angles such that

$\begin{array}{l}\angle DCA\cong \angle CAB;\\ \angle DAC\cong \angle ACB\end{array}$

Consider $\Delta ADC$and $\Delta ABC$, where $\angle DCA\cong \angle CAB$, $\angle DAC\cong \angle ACB$and $\overline{AC}$is common side in both the triangles therefore by ASA postulate, $\Delta ADC\cong \Delta ABC$.

As $\Delta ADC\cong \Delta ABC$, the angles of congruent triangles are congruent, that is, $\angle B\cong \angle D$.

As $\angle A,\angle C$and $\angle B,\angle D$are opposite angles of a parallelogram.

From $\angle A\cong \angle C$and $\angle B\cong \angle D$it can be concluded that the opposite angles of a parallelogram are congruent.