Question

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) ΔAPB ≅ ΔAQB
(ii) BP = BQ, i.e., B is equidistant from the arms of ∠A.

Answer 1

Given: In the given figure, ∠BAQ = ∠BAP, BP$\perp$AP and BQ$\perp$AQ.

To prove:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ, i.e., B is equidistant from the arms of ∠A.

Proof:
(i) In ΔAPB and ΔAQB,

∠BAQ = ∠BAP, (Given)
∠APB = ∠AQB = 90$°$ (Given, BP$\perp$AP and BQ$\perp$AQ)
AB = AB (Common side)

$\therefore$ By AAS congruence criteria,
ΔAPB ≅ ΔAQB

(ii)
$\because$ ΔAPB ≅ ΔAQB [From (i)]
$\therefore$ BP = BQ (CPCT)
Hence, B is equidistant from the arms of ∠A.