Question

line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendicular from B to the arms of angle A show that : ΔAPB =ΔAQB
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Answer 1

Answer:

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 }Tobeproved : BP = BQ

Proof :

As in (i) ΔAPB \cong≅ ΔAQB (proved)

so,

BP = BQ [ by CPCT ]

Hence, proved.

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

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

Step-by-step explanation:

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