Let $2a,2b,2c$ be the angles of the triangle.Quick angle chasing gives $\angle MA_1C=270-2a-3c$ AND $\angle EIC=270-2a-3c$. Now $\frac{NA_1}{A_1D}=\frac{A_1C}{A_1D}=\frac{IA_1}{IC}=\frac{IE}{IC}$. So $\triangle MA_1C~\triangle EIC$. So $C$ is the center of the spiral similarity that sends $A_1$ to $I$ and $M$ to $E$. Thus, $\angle EMC=\angle IA_1C$.

Alternatively, one could observe that $EN$ and $EC$ are isogonal w.r.t. $\angle B_1EA_1$. Therefore, we have that $\angle CEM = \angle CEA_1 + \angle MEA_1 = \angle CEA_1 + \angle NEA_1 = \angle B_1EA_1,$ and so it suffices to show that $EM = EC \cdot \cos \angle B_1EA_1.$ However, it's well-known that $EN = EC \cdot \cos \angle B_1EA_1$, and so we're done. $\square$