Question

In the figure given below, $E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $△ABE\sim △CFB$.

Answer 1

In ${△}^{\prime }sABE$ and $CFB$, we have

$\angle AEB=\angle CBF$ [Alternate angles]

$\angle A=\angle C$ [Opposite angles of a parallelogram]

Thus, by AA-criterion of similarity, we have

$△ABE\sim △CFB$. $\left[Henceproved\right]$