Question

help me fast............​

Answer 1

Step-by-step explanation:

In △s APB and ABQ, we have

∠APB=∠AQB (Each 90 )

∠PAB=∠QAB (AB bisect ∠PAQ)

AB=BA (common)

Therefore, △APB≅△ABQ (AAS)

⇒ BP=BQ (cpct)

Hence, B is equidistant from the arms of ∠A.

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