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 congruent to triangle AQB 2. BP equals to BQ or B is equidistant from the arms of angle A

Answer 1

$\mathbb{ SOLUTION }$ :

Given :
l is the bisector of ∠A
BP and BQ are perpendiculars from B to the arms of ∠A

(i) Now,
in ΔAPB and ΔAQR
∠QAB = ∠BAP [ ∴ l is the bisector of ∠A ]
AB = AB [ common side ]
∠APB = ∠AQB [ each angle = 90° ]

Hence,
by AAS congruence condition
ΔAPB $\cong$ ΔAQB

(ii) $\bf{ To \: be \: proved }$ : BP = BQ
Proof :
As in (i) ΔAPB $\cong$ ΔAQB (proved)
so,
BP = BQ [ by CPCT ]

Hence, proved.

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$\bf{AAS }$ : Angle-Angle-Side

$\bf{CPCT }$ : Corresponding Parts of Congruent Triangles

Answer 2

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