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
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