In the given figure, line l is the bisector of an angle ∠A and B is any point on l.

Question:

In the given figure, line l is the bisector of an angle ∠A and 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.

 

Solution:

Given: In the given figure, ∠BAQ = ∠BAPBP">AP and BQ">AQ.

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°">°°          (Given, BP">AP and BQ">AQ)
AB = AB                               (Common side)

"> By AAS congruence criteria,
ΔAPB ≅ ΔAQB

(ii)
"> ΔAPB ≅ ΔAQB            [From (i)]
"> BP = BQ                       (CPCT)
Hence, is equidistant from the arms of ∠A.

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