Question

The diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point on the side BC. Prove that DF × EF = FB × FA.

Answer 1

In $△AFDand△BFE$, we have
$\angle 1=\angle 2$ [Vertically opposite angles]
$\angle 3=\angle 4$ [Alternate angles]

So, by AA-criterion of similarity, we have
$△FBE\sim △FDA$
$\Rightarrow \frac{FB}{FD}=\frac{FD}{FA}$
$\Rightarrow \frac{FB}{DF}=\frac{EF}{FA}$
$DF\times EF=FB\times FA$