91Ó°ÊÓ

Consider the figure.

Consider the trapezoid $ABXY$with $BXâ‰\dots AY$.

Draw the line $BPâŠ\yen XY$and the line $AQâŠ\yen XY$at P and Q respectively.

In the triangles $\mathrm{Δ}BPX$and $\mathrm{Δ}AQY$,

$BXâ‰\dots AY$(Congruent sides of the trapezoid)

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

$∠BPXâ‰\dots ∠AQY\left(BPâŠ\yen XY,AQâŠ\yen XY\right)$

Therefore, by the right side, the hypotenuse and one side property the triangles $\mathrm{Δ}BPX$and$\mathrm{Δ}AQY$ are congruent.

Since the corresponding parts of the congruent triangles are equal such that $∠BPXâ‰\dots ∠AYQ$.

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

As the corresponding parts of congruent triangles are equal thus $∠XBPâ‰\dots ∠YAQ$.

Therefore, $∠Xâ‰\dots ∠Y,∠Bâ‰\dots ∠A$.

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