Question

5. Linel is the bisector of an angle Z A and B is any
point on I. BP and BQ are perpendiculars from B
to the arms of ZA (see Fig. 7.20). Show that:
(1) A APB = A AQB
(ii) BP = BQ or B is equidistant from the arms
В B.
of A.
Fig. 7.20​

Answer 1

Answer:

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 anus of ∠A.