Any solution ???

What am I missing here? $S(CFL) = \frac{S(FED) + S(CED)}{2} = S(FEL) + S(CLD)$ so the given sum is half the area of the hexagon irregardless of the given conditions.

It's $S(CFL) =$ $ \frac{S(FEC) + S(CFD)}{2} =$ $ S(CDEF)- \frac{S(FED) + S(CED)}{2} =$ $ S(CDEF) - (S(FEL) + S(CLD))$, so not quite what you assumed.

An observation: Let $X$ and $Y$ by the feet of the perpendiculars from $F$ and $C$ respectively onto $AB$. Similarly, let $Z$ and $W$ be the feet of the perpendiculars from $F$ and $C$ onto $DE$. Then: $\frac{BC}{CD}=\frac{EF}{FA}$ is equivalent to $FX \cdot CY = FZ \cdot CW$ which is equivalent to $X, Y, W, Z$ being concyclic (in that case note that $M$, the middle of $FC$ is the center of the circle). I have no idea how to relate this to the condition involving the area. In particular, I am not sure how to get something meaningful about the perpendiculars onto $FC$... Did anyone solve it?

See attachment below for a proof with no words which I hope all of you could understand. The diagram is drawn by my friend, not me.

Attachments:

Here are the details to accompany XmL's diagram. Let $C_1$ be the reflection of $C$ through $K$ and $C_2$ be the reflection of $C$ through $L$. Then, since $AK=KB$ and $EL=LD$, the quadrilaterals $C_1ACB$ and $EC_2DC$ are parallelograms. Then $\angle C_1AB=\angle CBA=\angle FEL$ and $\angle KAF=\angle EDC=\angle DEC_2$. Consequently, $\angle C_1AF=\angle FEC_2$. Since $C_1K=KC$, we have that $|KFC|=|C_1FK|$ and $|KBC|=|KAC_1|$. Thus $|KFC|-|FAK|-|KBC|=|AFC_1|$ and similarly $|FLC|-|FEL|-|LDC|=|FEC_2|$. The sum $|FAK|+|KCB|+|CFL|$ is half the area of the hexagon if and only if $|KFC|-|FAK|-|KBC|=|FLC|-|FEL|-|LDC|$ if and only if $|AFC_1|=|FEC_2|$ if and only if $C_1A\cdot AF=FE\cdot EC_2$ (by the sine area law, since $\angle C_1AF=\angle FEC_2$) if and only if $FA\cdot BC=FE\cdot CD$ if and only if $\frac{BC}{CD}=\frac{EF}{FA}$.

nice solution: but my solution some other. Let $M$ is midpoint $CF$. We have that $ [AKF]+[BKC]+[CLF]=[AMB]+[CMD]+[EMF] $ and we prove that $ [AMB]+[CMD]+[EMF]=[BMC]+[DME]+[FMA] $. Let $ B_1 $ and $ E_1 $ are reflection of the points $B$ and $E$ to the $M$. So $ [AFB_1]=[AMF]+[BMC]-[AMB] $ and $ [DCE_1]=[DMC]+[EMF]-[DME] $. Since $ \angle AFB_1=\angle DCE_1 $ WE GET EASILY THAT THING WHICH BE WE NEED PROOF!