Solution:

Triangles $EAB$ and $EDC$ are similar, and so are triangles $EBN$ and $ECQ$. Therefore, in directed angles it holds that $\text{MFE= MNE= BEN= QEC= EQM}$. Analogously it holds that $\text{QFE= EMQ}$, from which it follows that $F$ is the orthocenter of triangle $EMQ$. Hence, $EF \perp QM \| AC$.

Second solution. Consider the translation $\mathcal{T}$ by the vector $\frac{1}{2} \overrightarrow{AC}$. It holds that $\mathcal{T}(M)=$

$Q$ and $\mathcal{T}(N)=P$; denote $\mathcal{T}(E)=E^{\prime}$. From the similarity of triangles $AED$ and $BEC$ it follows that $\triangle AEM \sim \triangle BEP$, whence $\text{QE E= EMQ= MEA= BEP= QPE}$, so point $E^{\prime}$ lies on the circle $PEQ$. Therefore, the translation $\mathcal{T}$ maps the circle $MEN$ into the circle $PEQ$, so the line joining their centers is parallel to $AC$, from which the claim follows.