In $Fig.$,line segment $DF$ intersects the side $AC$ of a triangle $ABC$ at the point $E$ such that $E$ is the mid-point of $CA$ and $\angle AEF = \angle AFE$. Prove that
$\frac{BD}{CD} = \frac{BF}{CE}$

(N/A) Given: In $\triangle ABC$,$E$ is the mid-point of $CA$ and $\angle AEF = \angle AFE$.
To prove: $\frac{BD}{CD} = \frac{BF}{CE}$.
Construction: Draw a line $CG$ parallel to $EF$ such that $G$ lies on $AB$.
Proof: Since $E$ is the mid-point of $CA$,we have $CE = AE$ $(i)$.
In $\triangle ACG$,since $CG \parallel EF$ and $E$ is the mid-point of $CA$,by the converse of the Mid-point Theorem,$F$ is the mid-point of $AG$. Thus,$GF = AF$ $(ii)$.
Also,in $\triangle ACG$,since $CG \parallel EF$,by the Mid-point Theorem,$EF = \frac{1}{2} CG$. However,this is not needed here. Instead,consider $\triangle BDF$ and $\triangle BCG$. Since $CG \parallel EF$,by the Basic Proportionality Theorem in $\triangle BDF$,we have $\frac{BD}{CD} = \frac{BF}{GF}$.
Since $\angle AEF = \angle AFE$,in $\triangle AEF$,we have $AE = AF$. From $(i)$ and $(ii)$,$CE = AE = AF = GF$. Therefore,$CE = GF$.
Substituting $GF = CE$ in the ratio $\frac{BD}{CD} = \frac{BF}{GF}$,we get $\frac{BD}{CD} = \frac{BF}{CE}$.
Hence proved.