<s>[INST] 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.
 [/INST] 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 $\angle{CDE}=60°$.
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