In the given figure, line l is the bisector of an angle ∠ A and B is

1.	In the given figure, line l is the bisector of an angle ∠ A and B is any point on l. If BP and BQ are perpendiculars from B to the arms of ∠ A, show that(i) △ APB ≅ △ AQB(ii) BP = BQ, i.e., B is equidistant from the arms of ∠ A.
Answer» (i) Considering △ APB and △ AQB We know that ∠ APB = ∠ AQC = 90^o From the figure we know that l is the bisector of ∠ A So we get ∠ BAP = ∠ BAQ We know that AB is common i.e. AB = AB Therefore, by AAS congruence criterion we get △ APB ≅ △ AQB (ii) We know that △ APB ≅ △ AQB So it is proved that BP = BQ (c. p. c. t)

In the given figure, line l is the bisector of an angle ∠ A and B is any point on l. If BP and BQ are perpendiculars from B to the arms of ∠ A, show that(i) △ APB ≅ △ AQB(ii) BP = BQ, i.e., B is equidistant from the arms of ∠ A.

Answer»

(i) Considering △ APB and △ AQB

We know that

∠ APB = ∠ AQC = 90^o

From the figure we know that l is the bisector of ∠ A

So we get

∠ BAP = ∠ BAQ

We know that AB is common i.e. AB = AB

Therefore, by AAS congruence criterion we get

△ APB ≅ △ AQB

(ii) We know that △ APB ≅ △ AQB

So it is proved that BP = BQ (c. p. c. t)