Question

Clarify................

Answer 1

Taking FB = FB won't help as FB is a side of CFB but not a side of AEB.

To prove similarity:

1. CD || AB

So if you consider the transversal EB,

Angle CFB = Angle ABE

2. AE || BC and CD || AB

So ABCD is a parallelogram.

Angle BCF = Angle BAE

∆BCF ~ ∆EAB (AA similarity)