Question

Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:

1. ΔAPB ≅ ΔAQB
2. BP = BQ or B is equidistant from the arms of ∠A.

Answer 1

We have, l is the bisector of ∠QAP.

∴ ∠QAB = ∠PAB

and ∠Q = ∠P       ...[Each 90°]

⇒ ∠ABQ = ∠ABP    ...[By angle sum property of △]

i. Now, in △APB and △AQB, we have

∠ABP = ∠ABQ       ...[Proved above]

AB = BA            ...[Common]

∠PAB ≅ ∠QAB      ...[Given]

△APB ≅ △AQB     ...[By ASA congruency]

ii. Since, △APB ≅ △AQB

⇒ BP =  BQ     ...[By Corresponding parts of congruent triangles]

i.e., [Perpendicular distance of B from AP] = [Perpendicular distance of B from AQ]

Thus, the point B is equidistant from the arms of ∠A.