Develop a transformational proof that the base angles of an isosceles triangle must be congruent.
[b][u]Proof:[/u][/b] Consider an isosceles triangle ABC. By definition, we know that [math]\overline{AB}\cong\overline{BC}[/math]. [br][br]First, construct the perpendicular bisector of AC, which we will call [math]l_1[/math]. Let Z be the point at which [math]l_1[/math] intersects AC. As consequence of triangle ABC being isosceles, we know that [math]l_1[/math] will pass through point [math]B[/math].[br][br]Then, reflect triangle ABC about [math]l_1[/math]. As result of a reflection being an isometry, we know that it preserves angle measurements and distances between points. When triangle ABC is reflected, notice that BC maps to AB because [math]\overline{AZ}\cong\overline{CZ}[/math] by definition. Consequently, note that angles [math]\angle A[/math] and [math]\angle C[/math] switch places. However, also note that [math]\angle B[/math] remains in its original position because all points along [math]l_1[/math] are fixed regardless of the reflection. Since the distances and angles in this triangle are preserved, we know that when ABC is reflected, the triangle formed is congruent to ABC. For clarity, note that since the two segments which define angle A map to the two segments which define angle C, we know that [math]\angle A\cong\angle C[/math].