Question

Que: 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 ∠A. Show that:

(i) ΔAPB ≅ ΔAQB

(ii) BP = BQ or B is equidistant from the arms of ∠A.

Answer 1

Answer:

Solution:

It is given that the line “l” is the bisector of angle ∠A and the line segments BP and BQ are perpendiculars drawn from l.

(i) ΔAPB and ΔAQB are similar by AAS congruency because;

• ∠P = ∠Q (both are right angles)

• AB = AB (common arm)

∠BAP = ∠BAQ (As line l is the bisector of angle A)

So, ΔAPB ≅ ΔAQB.

(ii) By the rule of CPCT, BP = BQ. So, we can say point B is equidistant from the arms of ∠A.

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Hope it will be helpful :)

Answer 2

Answer:

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