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 Fig. 7.20). Show that:
(i) ∆ APB ≅ ∆ AQB
(ii) BP = BQ or B is equidistant from the arms of ∠ A

Given: l is the bisector of an angle ∠A and BP ⊥ AP and BQ ⊥ AQ

To Prove: ΔAPB ≅ ΔAQB and BP = BQ

Proof:  (i) In ΔAPB and ΔAQB,

∠BAP = ∠BAQ (l is the angle bisector of ∠A)

∠APB = ∠AQB (Each 90°)

AB = AB (Common)

∴ ΔAPB ≅ ΔAQB (By AAS congruence rule)

ii) Since, ΔAPB ≅ ΔAQB

∴ BP = BQ (By CPCT)

Or, it can be said that point B is equidistant from the arms of ∠A.