Question

am waiting for your response ​

Answer 1

Answer:

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 ∠A.