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Question

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

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Solution

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)


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