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Question

ABCD is rhombus. If AB is produced at F and BA is produced at E such that AB=AE=BF, then result will be ?

A
ED||CF
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B
ED>CF
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C
EDCF
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D
ED2+CF2=EF2
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Solution

The correct option is C EDCF

Produce EC and FC to meet at a point M.
Let each side of the rhombus be x. Then,
AB=BC=CD=AD=BF=AE=x
DAB+CBA=180o
DAE+CBF=180o
In ADF,
EAD+AED+ADE=180o ------ ( 1 )
In BCF,
CBF+CFB+BCF=180o ----- ( 2 )
AED=ADE --- ( 3 ) [ Base angles of an equal sides are also equal ]
CFB=BCF ---- ( 4 ) [ Base angles of an equal sides are also equal ]
Adding equation ( 1 ) and ( 2 ),
(EAD+CBF)+(AED+ADE)+(CFB+BCF)=360o
180o+2AED+2CFB=360o [ From ( 3 ) and ( 4 ) ]
AED+CFB=90o.
EMF=90o
EMFM
EDCF

1264331_1037685_ans_f3a02b519aef4ad890e4f8b49bff4b46.jpeg

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