Find the angles of triangle $NPQ$. 
$ABC$ is a triangle. $ACM$ and $BCN$ are equilateral triangles where $M$ and $N$ are at the outside of the triangle. $P$ is center of $ACM$. $Q$ is midpoint of AB. Then find the angles of the triangle $NPQ$.

I need the solution using homothety. I have already solved the problem, but I have not been able to get the solution with homothety.
My solution:(in short)
Let's take point $R$ at $PQ$ line where $PQ=QR$. Triangles $APQ$ and $BQR$ are congruent. Also notice that triangles NCP and $NQR$ are congruent. Now it's not hard to see NPR is equilateral triangle. Thus answer is $30°,60°,90°$.

Let $D$ be a midpoint of $BC$. Since $$\angle PCN = \angle QDN = 90+\gamma$$ and $${PC \over QD}  = {CN\over DN} = {2\over \sqrt{3}}$$ we see that $\triangle PCN\sim \triangle QDN$, so the spiral similarity at $N$ takes $\triangle PCN$ to $\triangle QDN$. But this spiral similarity induces new spiral similarity which has also center at $N$ and takes $\triangle CDN$ to $\triangle PQN$ so they have same angles.

