Find angles of triangle $DEC$ $ABC$ is an equaliterial triangle. $AC∥MN$ and $M$ and $N$ lies at $AB$ and $BC$ respectively. $D$ is centroid of $MBN$, $E$ is a midpoint of $AN$. Then, find the angles of the triangle $DEC$.
First of all I am sure that it can be solved by rotation. Unfortunately, I haven't managed to process much. Let $AC⊥BF$ and $F∈AC$. $B$,$F$,$D$ are collinear and $AF$=$FC$. Since $AE$=$EN$. $EF∥NC$ and $NC$=$2EF$.
I am suspecting angles of $DEC$ are 30°,60°,90°. So I am now trying to prove $CDEF$ is circumcircle.
 A: 
Let $\vec {BM} = a \vec {BA}$ and $\vec {BN} = a \vec {BC}$ to ensure $MN || AC$. Given the centroid $D$ and the midpoint $E$, 
$$\vec {BD} = \frac a3(\vec{BC}+\vec {BA} ),\>\>\>\>\>\vec {AE}  = \frac12\vec {AN} = \frac a2 \vec {BC} -\frac12\vec {BA} $$Then,
\begin{align}
& \vec {BE}  = \vec {AE} - \vec {AB} = \frac12( \vec {BN} - \vec {BA})  - \vec {AB} 
= \frac a2 \vec {BC} +\frac12 \vec {BA} \\
& \vec {ED}  = \vec {BD} - \vec {BE} = \frac a3(\vec{BC}+\vec {BA}) - (\frac a2 \vec {BC} +\frac12 \vec {BA}) = (\frac a3  -\frac12)\vec{BA} -\frac a6 \vec{BC}  \\
&  \vec {EC}  = \vec {AC} - \vec {AE}   = (1-\frac a2) \vec {BC} - \frac12 \vec {BA } \\
\end{align}
With $\vec {BC} \cdot \vec {BA } =\frac12|BC||BA|$ for equalaterral triangle $ABC$
$$\vec {ED}\cdot \vec {EC} = \left( (\frac a3  -\frac12)\vec{BA} -\frac a6 \vec{BC} \right)\cdot \left( (1-\frac a2) \vec {BC} - \frac12 \vec {BA } \right) = 0$$
Thus, $\angle DEC = 90^\circ$.
A: Another way to solve it is by analytical geometry. Considering the figure of the preceding answer, setting the  vertices $B(0,0)$, $C(1,0)$, and $A(1/2,\sqrt{3}/2)$, and placing the point $N$ in $(k,0)$ (with $0 \leq k \leq 1$), we directy get that:


*

*the coordinates of $D$ are $\Big(k/2,k/(2 \sqrt{3})\Big)$;

*the coordinates of $E$, midpoint between $A$ and $N$,  are $\Big((2k+1)/4,\sqrt{3}/4)\Big)$.


Now we can find the equation of the $DE$ line by solving the system 
$\left\{
 \begin{array}{ll}
  \frac{k}{2 \sqrt{3}}=a \frac{ k}{2} + b\\
  \frac{\sqrt{3}}{4}= a \frac{2k+1}{4} +b 
  \end{array}
  \right.
$
whose solutions lead to the line $$y=\frac{3-2k}{\sqrt{3}}+\frac{ k(k-1)}{\sqrt{3}}$$
Similarly, we can find the equation of the $CE$ line by solving the system 
$\left\{
 \begin{array}{ll}
  0=a+ b\\
  \frac{\sqrt{3}}{4}= a \frac{2k+1}{4} +b 
  \end{array}
  \right.
$
whose solutions lead to the line $$y=-\frac{\sqrt{3}}{3-2k}+\frac{ \sqrt{3}}{3-2k}$$
Since the slopes are negative reciprocals, the two lines are perpendicular and $\angle{DEC}=90°$. Now we can note that
$$\overline{DE}=\sqrt{\Big((2k+1)/4-k/2\Big)^2 +\Big(\sqrt{3}/4- k/(2 \sqrt{3})\Big) ^2}\\=\frac{\sqrt{k^2-3k+3}}{2 \sqrt{3}}$$
and
$$\overline{CE}=\sqrt{\Big((2k+1)/4-1\Big)^2 +\Big(\sqrt{3}/4\Big) ^2}\\=\frac{\sqrt{k^2-3k+3}}{2}$$
Since $\overline{CE}= \sqrt{3}\cdot \overline{DE } $, we directly get $\angle{DCE}=30°$ and $\angle{CDE}=60°$.
