Prove the Following:

Question

Prove: The bisector of the angle opposite the base of an isosceles triangle bisects the base and is perpendicular to the base

Answer 1

Let ABC be your isosceles triangle. with AB and BC being the equal sides and BD be where the angle bisector meets the base.

The bisector will form two triangles, triangle ABD and CBD. Since the triangle is isosceles, the two base angles A and C are equal. Also the angle being bisected forms two corresponding equal angles between the two triangles. Finally, the shared side, BD is equal by the reflective property. Thus triangles AB and CBD are congruent by AAS.

By CPCTE the bases of the two triangles are equal (AD = CD) meaning that the angle bisector bisects the base as well. Also, by CPCTE the two angles formed by the intersection of the angle bisector to the base are equal.

Since the two angles that are formed by the intersection of the angle bisector to the base form a linear pair, they sum to 180 degrees. Since they are equal, they are both ninety degrees and right angles. Thus the angle bisector is also perpendicular to the base.

Answer 2

Consider an isosceles triangle ΔABC, with base BC, AB = AC, and AD as the bisector of ∠BAC.

The bisector AD creates two triangles ΔABD and ΔACD. ΔABD has sides AB, AD, and BD and ΔACD has sides AC, AD, and CD.

Since AB = AC and the triangles share side AD then the third side must be equal, BD = CD = BC/2.

Since the triangles are congruent and ∠BDC = 180° ⇒ ∠BDA = ∠CDA = 90°.