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Question

In the given figure, AB||DE and BD||EF.Then, DC2=?



A

CF×AC
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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 : In Δ ABC, AB||DE and BD||EF



To prove : DC2=CF×AC

Proof : In Δ BCD, EF||BD (Given)CFDC=CECB....(1)(By corollary of BPT)In ΔABC, DE||AB(Given)DCAC=CECB....(2)(By corollary of BPT)From (1) and (2), We getCFDC=DCAC CF×AC=DC×DC CF×AC=DC2

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