In triangle $ABC$, $BD$ is a median.  $CF$ intersects $BD$ at $E$ so that $\overline{BE}=\overline{ED}$.  Point $F$ is on $AB$.  Then, if $\overline{BF}=5$, $\overline{BA}$ equals:
$ \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ \text{none of these} $
1. Given that $BD$ is a median in triangle $ABC$, it means that $D$ is the midpoint of $AC$. Therefore, $\overline{AD} = \overline{DC}$.
2. $CF$ intersects $BD$ at $E$ such that $\overline{BE} = \overline{ED}$. This implies that $E$ is the midpoint of $BD$.
3. Point $F$ is on $AB$ and $\overline{BF} = 5$.
4. We need to find the length of $\overline{BA}$.

To solve this, we can use the properties of similar triangles and the fact that $BD$ is a median.

5. Draw a line from $D$ parallel to $AB$ and let it intersect $CF$ at $K$. Since $D$ is the midpoint of $AC$ and $E$ is the midpoint of $BD$, $K$ will be the midpoint of $CF$.

6. By the properties of parallel lines and transversals, $\triangle FBE \cong \triangle KDE$ by Angle-Angle-Side (AAS) congruence:
   - $\angle FBE = \angle KDE$ (corresponding angles)
   - $\angle FEB = \angle KED$ (vertical angles)
   - $\overline{BE} = \overline{ED}$ (given)

7. Since $\triangle FBE \cong \triangle KDE$, it follows that $\overline{BF} = \overline{DK} = 5$.

8. In $\triangle AFC$, since $D$ is the midpoint of $AC$ and $K$ is the midpoint of $CF$, we have:
   \[
   \frac{\overline{DK}}{\overline{AF}} = \frac{\overline{DC}}{\overline{AC}} = \frac{1}{2}
   \]
   Therefore, $\overline{AF} = 2 \cdot \overline{DK} = 2 \cdot 5 = 10$.

9. Since $F$ is on $AB$ and $\overline{BF} = 5$, we have:
   \[
   \overline{BA} = \overline{BF} + \overline{FA} = 5 + 10 = 15
   \]

The final answer is $\boxed{15}$.