$E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $ riangle ABE \sim riangle CFB$.

Question

$E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $	riangle ABE \sim 	riangle CFB$.

Accepted Answer

Step 1: **Given:** $ABCD$ is a parallelogram. $E$ is a point on $AD$ produced. $BE$ intersects $CD$ at $F$.
**To prove:** $	riangle ABE \sim 	riangle CFB$.
Step 2: In $	riangle ABE$ and $	riangle CFB$: $\angle EAB = \angle BCF$ (opposite angles of parallelogram / $AD \parallel BC$, corresponding angles)
Step 3: $\angle AEB = \angle CBF$ (alternate interior angles, $AD \parallel BC$)
Step 4: $	herefore 	riangle ABE \sim 	riangle CFB$ (AA similarity)

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