Find the length of the line segment $CE$

Question

Let $BD$ be a median in $\triangle ABC$. The points $E,F$ trisect $BD$ such that $BE=EF=FD$. If $AB=1$ and $AF=AD$, find the length of the line segment $CE$.

Answer 1

Since $BF = DE$ and $DC = DA = AF$ and $$\angle CDE = 180-\angle ADF = 180-\angle AFD = \angle AFB $$ we have $$\Delta AFB \cong \Delta EDC\; (s.a.s)$$ so $CE = AB =1$

Answer 2

Hint: Note that $\angle AFD \cong \angle FDA$ because $m(AF) = m(DA)$ in $\triangle AFD$. Hence $\angle AFB \cong \angle EDC$ because they're complementary to congruent angles.

Also, note that $m(DA) = m(DC)$ because $BD$ is a median. Hence, $m(AF) = m(DC)$.

Now, what can you say about triangles $\triangle AFB$ and $\triangle EDC$ ?