Question

$ABCD$ is a parallelogram and $E$ is a point on $BC$. If the diagonal intersects $AE$ at $F$, prove that $AF\times FB=EF\times FD$.

Answer 1

Given:

In parallelogram $ABCD$, $E$ is any point on the side $BC$.

$AE$ meets the diagonal $BD$ at $F$.

To Prove: $AF\times FB=EF\times FD$

Proof:

Consider triangles $AFD$ and $BEF$,

$\angle ADB$ and $\angle CBD$ are alternate angles.

So, $\angle ADF=\angle EBF$ ......... (1)

and $\angle DFA=\angle BFE$ [Vertically opposite angles]

∴ By $A.A$– Similarity criterion,

$△AFD\sim △EFB$

Since, in two similar triangles, corresponding sides are in the same ratio.

$\Rightarrow \frac{AF}{FD}=\frac{EF}{FB}$

$\Rightarrow AF\times FB=EF\times FD$

Hence, proved.