Question

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.

Answer 1

We have:
$\angle AFD=\angle EFB(\mathrm{Vertically}\mathrm{Opposite}\mathrm{angles})$
$\because$ DA $\parallel$ BC
$\therefore$ $\angle DAF=\angle BEF$ (Alternate angles)
$△DAF~△BEF$ (AA similarity theorem)
$\Rightarrow \frac{AF}{EF}=\frac{FD}{FB}$
or, $AF\times FB=FD\times EF$
This completes the proof.