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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.
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Solution
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.
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