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Question

ABCD is a quadrilateral such that DAAB and CBAB, x = y and EF || AD such that EF bisects DEC. Then CD2 =

A
2CE2
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B
2AE2
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C
2CB2
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D
2AB2
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Solution

The correct option is A 2CE2
Given,
ABCD is a quadrilateral with DAAB and CBAB
and EF || AD
EF||BC
EFAB
Now, in ΔDAE and ΔCBE,
DAE=CBE [each 90]
Also, DEA=CEB
ΔDAEΔCBE [by AA similarity]
DEAE=CEBEDE=CE ...(i)
Now, in ΔDEC, right angled at E,
DE2+CE2=DC22DE2=2CE2=DC2

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