Question

8. 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
angle A, show that
(1) ΔΑΡΒ=~ΔΑQB
(ii) BP = BQ, i.e., B is equidistant from the
arms of angle A​

Answer 1

Given :

L is the bisector of angle A

=> Angle PAB = Angle BAQ

Angle P and Angle Q are 90°

Proof :

$Consider \: ΔΑΡΒ \: and \: ΔΑQB$

Angle P = Angle Q = 90° (Given)

Angle PAB = Angle BAQ (Bisectors)

AB = AB (Common)

Therefore by ASA Congruency, ΔΑΡΒ=~ΔΑQB

=> By CPCT, BP = BQ, i.e., B is equidistant from the arms of angle A

Hence Proved....

Mark as Brainliest....