Question
ABCD is a parallelogram and E is a point on BC. If the diagonal BD intersect AE at F, prove that AF × FB = EF × FD.
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Answer
Given: ABCD is a parallelogram and E is point on BC. Diagonals DB intersects AE at F.
To Prove: AF × FB = EF × FD
Proof: In $\triangle\text{AFD}$ and $\triangle\text{EFD}$
$\angle\text{AFD}=\angle\text{EFB}$ $($vertically opposite $\angle\text{s})$
$\angle\text{DAF}=\angle\text{BEF}$ $($ Alternate $\angle\text{s})$
$\therefore\triangle\text{AFD}\approx\triangle\text{EFD}$ [By AAA similarity]
$\therefore\frac{\text{AF}}{\text{EF}}=\frac{\text{FD}}{\text{FB}}$
$\text{AF}\times\text{FB}=\text{EF}\times\text{FD}$
Hence proved.
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