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.