Question

In the given figure, ABCD is a parallelogram, E is a point on BC and the diagonal BD intersects AE at F. Prove that : DF X FE=FB x FA.
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Answer 1

Given that, the diagonal BO of parallelogram ABCD intersects the segment AE at F, where E is any point on BC.

To prove: DF×EF=FB×FA

Based on the given information, we can draw the figure shown above.

Now, in △AFD and △BFE,

⇒ ∠FAD=∠FEB [ Alternate angles ]

⇒ ∠AFD=∠BFE [ Vertically opposite angles ]

∴ △ADF∼△BFE [ By AA similarity ]

∴ FADF= EFFB

∴ DF×EF=FB×FA[Hence proved]