In the given figure, line l is the bisector of an angle ∠A and B is any point on l.

Given: In the given figure, ∠BAQ = ∠BAP, BP⊥">⊥⊥AP and BQ⊥">⊥⊥AQ.

To prove:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ, i.e., B is equidistant from the arms of ∠A.

Proof:
(i) In ΔAPB and ΔAQB,

∠BAQ = ∠BAP,                   (Given)
∠APB = ∠AQB = 90°">°°          (Given, BP⊥">⊥⊥AP and BQ⊥">⊥⊥AQ)
AB = AB                               (Common side)

∴">∴∴ By AAS congruence criteria,
ΔAPB ≅ ΔAQB

(ii)
∵">∵∵ ΔAPB ≅ ΔAQB            [From (i)]
∴">∴∴ BP = BQ                       (CPCT)
Hence, B is equidistant from the arms of ∠A.