$E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $\Delta ABE \sim \Delta CFB$.
Step-by-Step Solution
Key Concept: In $\Delta ABE$ and $\Delta CFB$: $\angle A = \angle C$ (opposite angles of parallelogram) and $\angle AEB = \angle CBF$ (alternate interior angles for $AE \parallel BC$).
In $\Delta ABE$ and $\Delta CFB$:
1. $\angle A = \angle C$ (Opposite angles of parallelogram $ABCD$). [1.0 Mark]
2. $\angle AEB = \angle CBF$ (Alternate interior angles as $AE \parallel BC$ and $BE$ is transversal). [0.5 Mark]
By AA similarity criterion, $\Delta ABE \sim \Delta CFB$. Proved! [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Identifying $\angle A = \angle C$ in parallelogram: 1.0 Mark
Identifying alternate interior angles $\angle AEB = \angle CBF$: 0.5 Mark
Applying AA similarity criterion: 0.5 Mark
Correct Answer: