Let $ABC$ be an equilateral triangle and $D$ and $E$ two points on $[AB]$ such that $AD = DE = EB$. Let $F$ be a point on $BC$ such that $CF = AD$. Find the value of $\widehat{CDF} + \widehat{CEF}$.
The lines $D F$ and $A C$ are parallel according to Thales' theorem, so $\widehat{C D F}=\widehat{D C A}$. By symmetry of the equilateral triangle with respect to the midpoint of $[A B]$, $\widehat{D C A}=\widehat{B C E}$.

Thus, $\widehat{C D F}+\widehat{C E F}=\widehat{F C E}+\widehat{C E F}=180^{\circ}-\widehat{E F C}=180^{\circ}-\left(180^{\circ}-\widehat{B F E}\right)=\widehat{B F E}$. Since $B D F$ is equilateral (similar to $A B C$ and $E$ is the midpoint of $[B D]$, so $\widehat{B F E}=$ $\widehat{D F E}=\frac{60}{2}=30$. Therefore, $\widehat{C D F}+\widehat{C E F}=30$.