Question

In given figure 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 $DF\times EF=FB\times FA$

Answer 1

In $△AFD$ and $△BFE$, we have

$\angle 1=\angle 2$ [Vertically opposite angles]

$\angle 3=\angle 4$ [Alternate angles]

So, by AA-criterion of similarity,we have

$△FBE\sim △FDA$

$\Rightarrow$ $\frac{FB}{FD}=\frac{FE}{FA}$

$\Rightarrow$ $\frac{FB}{DF}=\frac{FE}{FA}$

$\Rightarrow$ $DF\times EF=FB\times FA$ $\left[Henceproved\right]$