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Question:

Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A. Show that: (i) △ABP ≅ △AQB (ii) BP = BQ or B is equidistant from the arms of ∠A.

Solution:

In △APB and △AQB
∠APB = ∠AQB (Each 90°)
∠PAB = ∠QAB (l is the angle bisector of ∠A)
AB = AB (Common)
∴ △APB ≅ △AQB (By AAS congruence rule)
∴ BP = BQ (By CPCT)
It can be said that B is equidistant from the arms of ∠A.