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.