Bisected median

Question

In triangle $ABC$, let $D$ be the foot of the median from $C$. Let $F$ be the foot of the segment from $B$ such that $BF$ bisects $CD$, and $E$ be their point of intersection. If $CF=12$, find $AF$.

I used mass point geometry here. I let $AF = x$ and the mass of $C$ be $1$, then the mass of $A$ must be $\frac{12}{x}$, and so is point $B$. Adding the masses we get that point $D$ has mass $\frac{24}{x}$. Since $CD$ is bisected, the value of $x$ must be $24$. Is this correct?

Answer 1

If $F\in AC$ your solution is right.

I like the following way.

Let $K\in AF$ such that $DK||BF.$

Thus, $$AK=KF=FC,$$ which gives $$AF=12+12=24.$$

Answer 2

Let $S$ denote areas. Since CE = ED, We have $S_{CFB}=S_{DFB}$. Then, $$1=\frac{S_{DFB}}{S_{CFB}} =\frac{\frac12 S_{ABF}}{\frac{CF}{AC}S_{ABC} } =\frac{\frac12 \frac{AF}{AC}S_{ABC}}{\frac{CF}{AC}S_{ABC} } =\frac{AF}{2CF}$$

Thus, AF = 2CF = 24.