Proof of parallel lines 
The quadrilateral ABCD is inscribed in circle W. F is the intersection point of AC and BD.  BA and CD meet at E. Let the projection of F on AB and CD be G and H, respectively. Let M and N be the midpoints of BC and EF, respectively. If the circumcircle of triangle MNH only meets segment CF at Q, and the circumcircle of triangle MNG only meets segment BF at P, prove that PQ is parallel to BC.

I do not know where to begin.


Possibly the last steps of a proof
This is no full proof, just some observations which might get you started, but which just as well might be leading in the completely wrong direction.
You could start from the end, i.e. with the last step of your proof, and work backwards.
$P$ is the midpoint of $FB$ and $Q$ is the midpoint of $FC$. Therefore the triangle $BCD$ is similar to $PQF$, and since they have the edges at $F$ in common, the edges opposite $F$ have to be parallel. So your next question is: why are these the midpoints?
You can observe that $NP$ is parallel to $AB=AE$, and $NQ$ is parallel to $CD=CE$. Since $N$ is the midpoint of $FE$, the $\triangle FNP$ is the result of dilating $\triangle FEB$ by a factor of $2$ with center $F$. Likewise for $\triangle FNQ$ and $\triangle FEC$. So this explains why $P$ and $Q$ are midpoints as observed, but leaves the question as to why these lines are parallel.
Bits and pieces
I don't have the answer to that question yet. But I have a few other observations which I have not proven either but which might be useful as piezes of this puzzle.


*

*$\measuredangle DBE = \measuredangle ECA = \measuredangle NMG = \measuredangle HMN$. The first equality is due to the cocircularity of $ABCD$, but the others are unexplained so far.

*$\measuredangle MGN = \measuredangle NHM$, which implies that the circles $MGN$ and $MHN$ have equal radius, and the triangles formed by these three points each are congruent.

