In a $\triangle{ABC}$, $D$ is the midpoint of $BC$ and $E$ is any point on $AC$. $F$ is a point on $BC$ such that $AF \parallel DE$. if $Ar(ABC)=256 cm^2$ then find $Ar(CEF)$.
I solved this by taking a special case where $E$ was the midpoint of $AC$. Then $F$ became coincident with $B$ because of midpoint theorem. So I could easily find out the area using the fact that $CF$ (or $CB$) was a median. So $Ar(CEF)=128 cm^2$.
I know that this is the answer for all cases but I don't know how to generalize it. Please help me with this.
$$[\triangle ADC]=[\frac{\triangle ABC} {2}]=128$$
D is the midpoint of BC
$$[\triangle FDE]=[\triangle ADE]$$
(same base DE, equal heights - parallel lines)
$$[\triangle CEF]=[\triangle FDE]+[\triangle CDE]=[\triangle ADE]+[\triangle CDE]=[\triangle ADC]=128$$