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B
CF÷AC
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C
DF×EB
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D
DF÷EB
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Solution
The correct option is A CF×AC Given : InΔABC,AB||DEandBD||EF
To prove : DC2=CF×AC
Proof : InΔBCD,EF||BD(Given)CFDC=CECB....(1)(By corollary of BPT)InΔABC,DE||AB(Given)DCAC=CECB....(2)(By corollary of BPT)From(1)and(2),WegetCFDC=DCAC⇒CF×AC=DC×DC∴CF×AC=DC2