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) Δ APB ≅ Δ 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

(i) In Δ APB and Δ AQB,

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

⇒ ∠APB = ∠AQB (Each equal to 90°)

⇒ AB = AB (Common side)

∴ Δ APB ≅ Δ AQB (By A.A.S. congruence rule)

Hence, proved that Δ APB ≅ Δ AQB.

(ii) As,

Δ APB ≅ Δ AQB

We know that,

Corresponding parts of congruent triangles are equal.

∴ BP = BQ (By C.P.C.T.)

Hence, proved that BP = BQ or point B is equidistant from the arms of ∠A.