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 angle A. Show that :
1) Triangle APB is congruent to triangle AQB.
2) BP=BQ or B is equidistant from the arms of angle A

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.