Question

$E$ 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 $△ABE$ and $△CFB,$
$\angle ABE=\angle CFB$ (Alternate angles)
$\angle BAE=\angle BCF$ (opposite angles of a parallelogram)
$\therefore$ By AA criterion of similarity, $△ABE$ $\sim$ $△CFB$