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.

Answer 1

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.