Question

आकृती मध्ये $\Box ABCD$ समांतरभुज चौकोन आहे. किरण AB वर बिंदू E असा आहे की BE = AB. तर सिद्ध करा, की रेषा ED ही रेख BC ला F मध्ये दुभागते.

Answer 1

Solution :

Given :

ABCD is a parallelogram .

AB = BE (given)----( 1 )

To prove : F is the midpoint

of BC ( BF = FC )

Proof :

In ∆BEF and ∆DCF

BE = DC [ S ]

{ Since , AB = DC opposite

sides of a parallelogram.

And BE = DC from (1) }

<FBE = <FCD

[Alternate angles ]

<BEF = <CDF

[ Alternate angles ]

Therefore ,

∆BEF is congruent to ∆DCF

[ ASA congruence rule ]

BF = FC [ CPCT ]

F is the midpoint of BC.

Hence , prooved .

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