ABCD is a parallelogram and E …

Given: ABCD is a parallelogram and E is a point on BC. The diagonal BD intersects AE at F.
To prove: {tex}A F \times F B = E F \times F D{/tex}
Proof: Since ABCD is a parallelogram, then its opposite sides must be parallel.
{tex}\therefore{/tex}In {tex}\triangle ADF{/tex} and {tex}\triangle EBF{/tex}
{tex}\angle \mathrm { FDA } = \angle \mathrm { EBF }{/tex} and {tex}\angle F A D = \angle F E B{/tex} [Alternate interior angles]
{tex}\angle A F D = \angle B F E{/tex} [vertically opposite angles]
{tex}{/tex} Therefore,by AAA criteria of similar triangles,we have,
{tex}\triangle \mathrm { ADF } = \triangle \mathrm { EBF }{/tex}
Since  the corresponding sides of similar triangles are proportional. Therefore,we have,
{tex} \frac { A F } { F D } = \frac { E F } { F B }{/tex}
{tex}\Rightarrow A F \times F B = E F \times F D{/tex}