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.
