In parallelogram \(ABCD\), \(E\) lies on the extension of \(AD\), while \(BE\) meets \(CD\) at \(F\). Prove that \(\triangle ABE\) and \(\triangle CFB\) are similar.
A parallelogram \(ABCD\), where \(E\) is point on side \(AD\) produced and \(BE\) intersects \(CD\) at \(F\).
In parallelogram \(ABCD\), opposite angles are .
So, \( \angle A\ =\) \(\rightarrow (1) \)
Also, in parallelogram \(ABCD\), opposite sides are .
So, \(AD\ \parallel \)
Since \(AE\) is extended to \(AD\), \(AE \parallel BC\) and \(BE\) is the transversal.
\(\angle AEB\ =\) (Alternate angles) \(\rightarrow (2)\)
Now, in \(\triangle ABE\) and \(\triangle CFB\),
\(\angle A\ =\) (From (1))
\(\angle AEB\ =\) (From (2))
Thus, \(\triangle ABE \sim \triangle CFB\) by similarity criterion.
Hence proved.