The points $D$ and $E$ divide the side $AB$ of an equilateral triangle in three equal parts, such that $D$ is inbetween $A$ and $E$. The point $F$ is on $[BC]$ such that $CF=AD$. Calculate $\widehat{CDF}+\widehat{CEF}$
I've tried to apply the law of cosinus multiple times, but that seems like a bad proof... Does anyone have a better idea?
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We see that $\triangle{DFB}$ is an equilateral triangle and $\triangle{CDE}$ is an isosceles triangle.
Since $\angle{FEB}=90^\circ$, we get $$\angle{DCE}=180^\circ-2\angle{CED}=180^\circ-2(90^\circ-\angle{CEF})=2\angle{CEF}$$ So, letting $G$ be the intersection point of $CE$ with $DF$ where $\angle{GFE}=30^\circ$, we get $$\angle{GCD}+\angle{GDC}=\angle{GFE}+\angle{GEF},$$ from which we get $$\angle{CDF}+\angle{CEF}=\color{red}{30^\circ}$$
Choose point $G$ on $AC$ such that $CG = CF$. By the symmetry, $\widehat{CEF} = \widehat{CDG}$. So, the sum you're looking for is actually the angle $\widehat{GDF}$. At this point, you gotta realize $GF = AD$, and $GF$ is parallel to $AB$, thus, $ADFG$ is a parallelogram. As $\frac{AG}{AD} =2$, $\widehat{GDA} = 90^\circ$. Giving, $\widehat{GDF} = 30^\circ$.