#### Answer

$\overline{BA}$

#### Work Step by Step

$\overline{AF}$ is congruent to $\overline{FC}$, meaning the midpoint of $\overline{AC}$ occurs at point $F$; $\overline{BE}$ is congruent to $\overline{EC}$, meaning the midpoint of $\overline{BC}$ occurs at point $E$. Therefore, $\overline{EF}$ joins $\overline{BC}$ at its midpoint $E$ and $\overline{AC}$ at its midpoint $F$.
According to the triangle midsegment theorem, if a line segment joins two sides of a triangle at their midpoints, then that line segment is parallel to the third side of that triangle and is half as long as that third side.
Knowing this information, we can deduce that $\overline{BA}$, which is the third side, is parallel to $\overline{EF}$, which is the line segment.
$\overline{EF} \parallel \overline{BA}$