Problem :
Let $ABCD$ a square with $AB=1$ and $BMC،DNC$ equilateral triangles such $M$ inside the square and $N$ outside square
Prove that : $(BN)//(DM)$
My attempt :
take axes $(A,\vec{AB},\vec{AD})$
Then I'm going to find the coordinates of $N,M,B,D$
$B(0,1),D(1,0)$ now $M,N$
$x_{M}=\cos \frac{π}{3}=\frac{1}{2}$ , $y_{M}=1-\cos \frac{π}{6}=\frac{2-\sqrt{3}}{2}$
So : $M\left(\frac{1}{2},\frac{2-\sqrt{3}}{2}\right)$
Same method we find we obtaine :
$N\left(\frac{2+\sqrt{3}}{2},\frac{1}{2}\right)$
Now :
$m_{BN}=\frac{\frac{-\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}$
$m_{DM}=\frac{1}{\sqrt{3}}$
Conclusion : $m_{BN}.m_{DM}=-1$ this mean $(DM)//(BN)$
This is my way I need see different way for example by angles or similar triangle Homothetic transformation or rotational or barycenter for example
I have already to see your hints 
Let $P$ be the midpoint of $DM$ and let $Q$ be the intersection of $CP$ and $BN$. Let $E$ be the intersection of $BN$ and $CM$, and $F$ be that of $BN$ and $CD$.
Since $CM=CD$, it is clear that $CP\perp DM$ and $CP$ bisects $\angle MCD$.
Since $CB=CN$, we have $$\tag{1}\angle CBN=\angle CNB.$$ Since $\angle BCM=\angle DCN=60^\circ$, we have $$\tag{2}\angle ECN=\angle ECF+\angle DCN=\angle ECF+\angle BCM=\angle BCF.$$ $(1)$, $(2)$, and the fact that $BC=CN$, implies that $$\triangle BCF\text{ and }\triangle NCE\text{ are congruent},$$ from which we deduce that $$\tag{3} BF=NE.$$ Since $BF=BE+EF$ and $NE=NF+EF$, $(3)$ implies that $$\tag{4} BE=FH.$$ Using $(1)$, $(4)$, and the condition that $BC=CN$, one gets that $$\triangle BEC\text{ and }\triangle NFC\text{ are congruent},$$ which yields that $CE=CF$, and so $\triangle ECF$ is an isosceles triangle. Since $CP$ bisects $\angle ECF$, we see that $CQ\perp EF$.
Since both $DM$ and $NB$ are perpendicular to $CP$, it follows that $DM\parallel BN$. $\mathbf{Q.E.D.}$