证明:$ \because AE // BF $(已知),$ \therefore \angle AEC = \angle BFD $(两直线平行,内错角相等).在$ \triangle AEC $和$ \triangle BFD $中,$ \begin{cases} AE = BF \ (已知) , \\\angle AEC = \angle BFD \ (已证) , \\CE = DF \ (已知) , \end{cases} $$ \therefore \triangle AEC \cong \triangle BFD \ (SAS) $