In parallelogram ABCD, side AD is produced to a point E and BE intersects CD at F. Prove that $ riangle ABE \sim riangle CFB$.

Question

In parallelogram ABCD, side AD is produced to a point E and BE intersects CD at F. Prove that $	riangle ABE \sim 	riangle CFB$.

Accepted Answer

Step 1: Parallelogram ABCD with AD produced to E, BE intersecting CD at F
Step 2: In $	riangle ABE$ and $	riangle CFB$:
$\angle AEB = \angle CBF$ (common angle)
$\angle A = \angle C$ (opposite angles of parallelogram)
Step 3: $	herefore 	riangle ABE \sim 	riangle CFB$ (AA similarity)

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