Question

Fig. 7.19
Line l is the bisector of an angle A and B is any
point on I . BP and BQ are perpendiculars from B
to the arms of angle A Show that:
ΔΑΡΒ is congruent to AQB
BP = BQ or B is equidistant from the arms of A
​

Answer 1

Step-by-step explanation:

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