Prove that the base angles of an isosceles trapezoid are congruent.

Consider the figure.

Consider the trapezoid $ABXY$with $BX\cong AY$.

Draw the line $BP\perp XY$and the line $AQ\perp XY$at P and Q respectively.

In the triangles $\Delta BPX$and $\Delta AQY$,

$BX\cong AY$(Congruent sides of the trapezoid)

$\overline{BP}=\overline{AQ}$(distance between the parallel lines of the trapezoid)

$\angle BPX\cong \angle AQY\left(BP\perp XY,AQ\perp XY\right)$

Therefore, by the right side, the hypotenuse and one side property the triangles $\Delta BPX$and$\Delta AQY$ are congruent.

Since the corresponding parts of the congruent triangles are equal such that $\angle BPX\cong \angle AYQ$.

The interior angles on the same side of the transversal are supplementary such that $\angle Q+\angle P={180}^{\circ }$.

As the corresponding parts of congruent triangles are equal thus $\angle XBP\cong \angle YAQ$.

Therefore, $\angle X\cong \angle Y,\angle B\cong \angle A$.

Thus, the base angles of the isosceles trapezoid are congruent.