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