Consider an isosceles trapezoid $ABCD.$

The goal of this proof is to show that $\angle A\cong \angle B$ and $\angle C\cong \angle D.$

Proving That $\angle A\cong \angle B$

Start by drawing an auxiliary line that passes through $C$ and is parallel to $\overline{AD}.$ Let $E$ be the intersection point of that line and $\overline{AB}.$

Quadrilateral $AECD$ has two pairs of parallel sides, so by definition, $AECD$ is a parallelogram. Consequently, by the Parallelogram Opposite Sides Theorem, the sides opposite to each other are congruent. By the definition of an isosceles trapezoid, $\overline{AD}$ and $\overline{BC}$ are congruent. Therefore, by the Transitive Property of Congruence, $\overline{BC}$ and $\overline{EC}$ are also congruent. This implies that $△BEC$ is an isosceles triangle. Considering the Isosceles Triangle Theorem, it can be stated that $\angle BEC$ and $\angle B$ are congruent.

Additionally, $\angle A$ and $\angle BEC$ are congruent by the Corresponding Angles Theorem.

Therefore, using the Transitive Property of Congruence again, it is can be stated that $\angle A\cong \angle B.$

Proving That $\angle C\cong \angle D$

The bases of each trapezoid are parallel and its nonparallel sides can be considered transversals. This means that $\angle A$ and $\angle D,$ as well as $\angle B$ and $\angle C,$ are consecutive interior angles.

By the Consecutive Angles Theorem, these two pairs of angles are supplementary. In other words, their corresponding sums are $180^{\circ }.$ Since $\angle A$ and $\angle B$ are congruent, they have the same measure. Therefore, $m\angle B$ can be substituted for $m\angle A$ in Equation (I). Next, Equation (I) can be solved for $m\angle D$ and Equation (II) can be solved for $m\angle C.$ The right-hand sides of the equations are the same, so $\angle C$ and $\angle D$ have the same measure. Therefore, $\angle C$ and $\angle D$ are congruent angles. This concludes the proof.