In the figure, AB and CD are perpendiculars on BD. Also AB=CD and AF=CE. Prove that BE=FD.

In ∆ABF  and ∆CDE
∠B = ∠D = 90^o  {from figure}
Side AB = Side CD  {from figure}
Side AF = Side CE  {from figure}
Therefore, by RHS rule, ∆ABF and ∆CDE are concurrent.
By CPCD,
Side BF = Side DE
Subtracting common length EF from both the sides, we get
BF – EF = DE – EF
BE = FD
Hence Proved.