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. please don't spam or else you shall be reported.

In the given figure, ABCD is a parallelogram, E is a point on BC and t

1.	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. please don't spam or else you shall be reported.
Answer» hat, the diagonal BO of parallelogram ABCD intersects the segment AE at F, where E is any POINT on BC.To prove: DF×EF=FB×FABased 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]