Question

$\text{In the given figure, line l is the bisector of an angle}\angle A\text{and B is any point on l.}\text{If BP and BQ are perpendiculars from B to the arms of}\angle A,\text{show that}\left(i\right)\Delta APB\cong \Delta AQB\text{(ii) BP = BQ, i.e., B is equidistant from the armos of}\angle A.$

Answer 1

Given: In the given figure, ∠ BAQ = ∠ BAP , BP ⊥ AP and BQ ⊥ AQ.
To prove:
(i) Δ APB ≅ Δ AQB
(ii) BP = BQ, i.e., B is equidistant from the arms of ∠ A.
Proof:
(i) In Δ APB and Δ AQB,
∠ BAQ = ∠ BAP (Given) ,
∠ APB = ∠ AQB = 90° (Given, BP ⊥ AP and BQ ⊥ AQ)
AB = AB (Common side)

∴ By AAS congruence criteria,
Δ APB ≅ Δ AQB
(ii)
∵ Δ APB ≅ Δ AQB
∴ BP = BQ (CPCT)
Hence, B is equidistant from the arms of ∠ A.