Answer:
See Explanation
Step-by-step explanation:
[tex] AC \cong\DB... (given)\\\|
(AB + BC) \cong (DC + BC) \\\\
AB \cong DC.... (1)\\\\
In\: \triangle ABC \: \&\: \triangle DCE\\
FA \cong ED... (given) \\
\angle A \cong \angle D.... (given) \\
AB \cong DC....[From\: (1)] \\
\therefore \triangle ABC \: \cong\: \triangle DCE.. (by\: SAS\: postulate) \\
\therefore BF \cong CE.. (csst) [/tex]
