The question is as follows in the given figure $F$ is the midpoint of $DC$ and $\angle CFE = \angle CEF$ prove that $$\frac{AB}{BD} = \frac{AE}{FD}$$
The figure is here
According to me the question seems wrong but i cant find a way to prove it could anyone prove the question or can someone disprove the question using contradiction.
This statement is correct.
Consider the triangles $\triangle ABE$ and $\triangle BDF$ (see the figure).
Let $\alpha=\angle CFE=\angle CEF$ and $\beta=\angle ABE$. By the law of sines we get: $$ \triangle ABE:\ \frac{AB}{\sin(180^\circ-\alpha)}=\frac{AE}{\sin\beta} $$ and $$ \triangle BDF:\ \frac{BD}{\sin\alpha}=\frac{FD}{\sin\beta}. $$
Since $\sin(180^\circ-\alpha)=\sin\alpha$, it follows that $$ \frac{AB}{BD}=\frac{AE}{FD}. $$