3B. In triangle $ABC$, a median $BD$ is drawn. Points $E$ and $F$ divide the median into three equal segments $\overline{BE}=\overline{EF}=\overline{FD}$. If $\overline{AB}=1$ and $\overline{AF}=\overline{AD}$, determine the length of segment $CE$.
Solution. Since $\overline{A F}=\overline{A D}$, triangle $D A F$ is isosceles, so $\measuredangle A D F = \measuredangle A F D$, and from this we have $\measuredangle B F A = \measuredangle C D E$. Segment $B D$ is a median, so $\overline{C D} = \overline{A F}$, and from the condition of the problem $\overline{B F} = \overline{D E}$. Therefore, by the $S A S$ criterion, triangles $C D E$ and $A F B$ are congruent, so $\overline{C E} = \overline{A B} = 1$.

![](https://cdn.mathpix.com/cropped/2024_06_05_a6c2bf676fcc39861820g-06.jpg?height=383&width=507&top_left_y=1015&top_left_x=958)

## III year