Question

If AP bisects angle BAC and M is any point on AP, prove that the perpendiculars drawn from M to AB and AC are equal.

Answer 1

From M, draw ML such that ML is perpendicular to AB and MN is perpendicular to AC

In ΔALM and ΔANM

∠LAM = ∠MAN             ...[∵ AP is the bisector of BAC]

∠ALM = ∠ANM = 90°   ...[∵ ML ⊥ AB, MN ⊥ AC]

AM = AM                   ...[Common]

∴ By angle-angle-Side criterion of congruence, 

ΔALM ≅ ΔANM

The corresponding parts of the congruent triangles are congruent.

∴ ML = MN               ...[c. p. c. t]

Hence, proved.