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In figure, DE || AB and BD || BEF. Prove...

In figure, `DE || AB and BD || BEF.` Prove that `DC^2=CF xx AC`

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In `DeltaABC` , since DE||AB
`(CD)/(CA)= (CE)/(CB)` (by B.P theorem) …(1)
In `DeltaCDB`since FE||DB,
`(CF)/(CD)=(CE)/(CB)` (by B.P theorem) …(2)
From (1) and (2) we have,
`(CD)/(CA)=(CF)/(CD)`
`CD^(2)=CFxxCA` Hence proved.
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