Question

please any one can say answer​

Answer 1

Answer:

Hii..!

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→ In △APB and △AQB

∠APB=∠AQB (Each 90⁰)

∠PAB=∠QAB (l is the angle bisector of ∠A)

AB=AB (Common)

∴△APB≅△AQB (By AAS congruence rule)

∴BP=BQ (By CPCT)

It can be said that B is equidistant from the arms of ∠A.

Hence Solved!

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Hope it helps you dear!

Answer 2

Answer:

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