Answer: C

Small Hint:

Use coordinates or masses, noting that $D$ is the midpoint of $AC$ and $E$ is the midpoint of $BD$

Big Hint:

Write $E$ both as $\frac{B+D}{2}$ and as a point on line $CF$

Solution:

Use vectors with $A=\mathbf0,\ B=\mathbf b,$ and $C=\mathbf c.$ Since $D=\frac{\mathbf c}{2}$ and $E$ is the midpoint of $BD,$ $$E=\frac12\mathbf b+\frac14\mathbf c.$$ A point $F$ on $AB$ has the form $F=t\mathbf b.$ Since $E$ lies on $CF,$ comparison of the $\mathbf c$-coefficient shows that $E=\tfrac14C+\tfrac34F.$ Hence $$\frac34t=\frac12,$$ so $t=\frac{2}{3}.$ Thus $\frac{AF}{AB}=\frac{2}{3},$ and $\frac{BF}{BA}=\frac{1}{3}.$ Since $BF=5,$ $BA=15.$

Therefore, the correct answer is C.