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 (see the given figure). Show that: (i) ΔAPB ≅ ΔAQB (ii) BP = BQ or B is equidistant from the arms of ∠A.

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Solution

## In ΔAPB and ΔAQB, ∠APB = ∠AQB (Each 90º) ∠PAB = ∠QAB (l is the angle bisector of ∠A) AB = AB (Common) ∴ ΔAPB ≅ ΔAQB (By AAS congruence rule) ∴ BP = BQ (By CPCT) Or, it can be said that B is equidistant from the arms of ∠A.

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