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Question

In the given figure, line l is the bisector of an angle A and B is any point on l.If BP and BQ are perpendiculars from B to the arms of A, show that(i) ΔAPB ΔAQB(ii) BP = BQ, i.e., B is equidistant from the armos of A.

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Solution

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.


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