ABCD is a parallelogram and E is a point on BC. If the diagonal BD intersects AE at F, prove that AF × FB = EF × FD.

Given that, AB DC & AD BC

To Prove: AF × FB = EF × FD


Proof: In ∆DAF & ∆BEF


DAF = BEF [ they are alternate angles]


AFD = EFB [ they are vertically opposite angles]


This implies that ∆DAF ∆BEF by AA-similarity criteria.



Now cross-multiply them,


AF × FB = FD × EF


Hence, proved.


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