Question

Please don't delete this help me through this​

Answer 1

Answer:

In △APB and △AQB

∠APB=∠AQB (Each 90°)

∠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.