Question

Show that the perpendiculars drawn from the vertices of the base of an isosceles triangle to the opposite sides are equal.

Answer 1

Given: AB = CA and ∠BQA = ∠CPA = 90°

To Prove: CP = BQ

Proof:

In ΔCPA and ΔBQA:

∠A = ∠A (Common)

AB = AC (Given)

∠BQA = ∠CPA (Given)

ΔCPA ≅ ΔBQA (AAS congruency)

∴ CP = BQ (Corresponding parts of congruent triangles)