Given: $ABCD$ is a parallelogram. $E$ is a point on $AD$ produced. $BE$ intersects $CD$ at $F$.

To Prove: $\Delta ABE \sim \Delta CFB$

Proof:

In parallelogram $ABCD$, $AB \parallel CD$ and $AD \parallel BC$ (opposite sides of parallelogram are parallel).

Also, opposite angles of a parallelogram are equal, so $\angle BAD = \angle BCD$.

Since $AB \parallel CD$ and $BE$ is a transversal intersecting them, $\angle ABE = \angle CFB$ (corresponding angles).

Now, since $E$ lies on $AD$ produced, $AE$ is along $AD$. Similarly, $F$ lies on $CD$, so $CF$ is along $CD$.

Therefore, $\angle BAE = \angle BAD$ and $\angle BCF = \angle BCD$.

But $\angle BAD = \angle BCD$ (opposite angles of parallelogram)

Hence, $\angle BAE = \angle BCF$

In $\Delta ABE$ and $\Delta CFB$:

• $\angle ABE = \angle CFB$ (proved above)
• $\angle BAE = \angle BCF$ (proved above)

Therefore, by AA similarity criterion, $\Delta ABE \sim \Delta CFB$.

Hence proved.