E is a point on the side AD (produced) of a parallelogram ABCD and BE intersects CD at F. Show that \(\triangle ABE \sim \triangle CFB\).
Given: ABCD is a parallelogram. E lies on AD produced beyond D and BE meets CD at F.
To prove: \(\triangle ABE \sim \triangle CFB\).
In a parallelogram, opposite sides are parallel:
\[AB \parallel CD, \quad AD \parallel BC.\]
Since E lies on AD produced, AE is a straight-line extension of AD and hence:
\[AE \parallel BC.\]
Angle \(\angle ABE\) is formed by BA and BE.
Angle \(\angle CFB\) is formed by FC and FB.
But FC lies on CD and \(CD \parallel AB\). Also, FB is common to both angles as a side.
Therefore, corresponding angles are equal (alternate interior angles with transversal BF):
\[\angle ABE = \angle CFB.\]
Angle \(\angle AEB\) is formed by AE and EB.
Angle \(\angle CBF\) is formed by CB and BF.
We have \(AE \parallel BC\) and \(BE\) is the same as \(BF\) along line B–E–F.
Thus, these two angles are also equal (corresponding/alternate interior angles):
\[\angle AEB = \angle CBF.\]
We have two pairs of equal angles:
\[\angle ABE = \angle CFB, \quad \angle AEB = \angle CBF.\]
Therefore, by the AA similarity criterion:
\[\triangle ABE \sim \triangle CFB.\]
Hence proved.