Question

line l is the bisector of an angle A and B is any point on L. BP and BQ are perpendicular from B to the arms of angle A show that

1. triangle APB is congruent to triangle AQB

2. show that BP = BQ or B is equidistant from the arms of angle A ​

Answer 1

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☯ SOLUTION:-

♨ Given:-

Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of angle A.

♨ To Prove:-

i) ∆APB is congruent to ∆AQB

ii) BP = BQ

♨ Proof:-

i) In ∆APB and ∆AQB,

angle BAP = angle BAQ [Since l is the bisector of angle ]

AB = AB [Common]

angle BPA = angle BQA [Each 90° and since BP and BQ are perpendiculars from B to the arms of angle A]

Therefore,

By AAS rule, ∆APB is congruent to ∆AQB.

ii) Now, since ∆APB is congruent to ∆AQB,

Therefore,

By C.P.C.T,

BP=BQ or B is equidistant from the arms of angle A.

HOPE IT HELPS UHH ♥ ✅