Saudi Arabian Mathematical Competitions

Let $K$ be the midpoint of $A B$ and $L$ the intersection point of $D K$ with $B E$.

Because $D$ and $K$ are midpoints of $B C$ and $B A$, respectively, segment $D K$ is parallel to side $A B$ and we have
$$\frac{E F}{L E}=\frac{E A}{D E}=1 .$$
Because $E$ and $K$ are midpoints of $A D$ and $A B$, respectively, the point $L$ is the centroid of triangle $A B D$. Therefore
$$\frac{L E}{B E}=\frac{1}{3} .$$
We deduce that the ratio of the areas of the triangles
$$\frac{[A E F]}{[A B E]}=\frac{E F}{B E}=\frac{1}{3} .$$
Therefore
$$\frac{[A E F]}{[A B C]}=\frac{[A E F]}{[A B E]} \cdot \frac{[A B E]}{[A B D]} \cdot \frac{[A B D]}{[A B C]}=\frac{1}{12},$$
since $D$ and $E$ are midpoints of $B C$ and $A D$, respectively. We deduce that
$$$$
$\text{[C}$ F E D]}{[A B $\text{C] = [A}$ D C]-[A E F]}{[A B ${C]}=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}$ .

Let
$$x=\frac{[C F D]}{[A B C]} \quad \text{ and } \quad y=\frac{[D F E]}{[A B C]}$$
be the ratios of the areas.
Because $E$ is the midpoint of $A D$, we have
$$[A B E]=[B D E]=\frac{1}{2}[A B D] \quad \text{ and } \quad[A E F]=[D F E] .$$
Because $D$ is the midpoint of $B C$, we have
$$[A B D]=[A D C]=\frac{1}{2}[A B C] \quad \text{ and } \quad[F B D]=[F D C] .$$

We deduce that
$$x+2 y=\frac{1}{2} \quad \text{ and } \quad x=y+\frac{1}{4},$$
and therefore
$$x=\frac{1}{3} \quad \text{ and } \quad y=\frac{1}{12} .$$
This proves that
$$$$
$\text{[C}$ F E D]}{[A B ${C]}=x+y=\frac{5}{12}$ .