Prove that in acute-angled triangle middle point of a sector lies at a certain line

Segments AD, BE are heights of acute-angled triangle ABC. Point M is middle point of AB segment. Points P, Q are symetrical to point M under respectively line AD, BE. Show that center point of DE segment lies on straight PQ.

Here's a picture for better refrance https://puu.sh/recsD.png (I've used geogebra to create a small applet). Points G and H are cross points of AD with PM and BE with MQ. Point F is middle point of DE (probably PQ as well)

I've been trying to prove it for past few hours without much progress. I figured that:

• PM || BC, QM || AC
• angle PMQ = angle ACB
• triangles CEB and CDA are simillar.

Any help appreciated :)

There are several variations, but let me mention one. Segment $MH$ is orthogonal to $BE$ and since $AE$ is orthogonal $BE$ the two segments $MH$ and $AE$ are parallel. Since $M$ is a midpoint of $AB$, segment $MH$ is half of $AE$ in length. Analogously, $MG$ is parallel to $BD$ and since $M$ is a midpoint of $AB$, point $G$ is a midpoint of $AD$.
Look at quadrilateral $MGFH$. Segment $FG$ is a midpoint-segment in triangle $AED$ and is therefore parallel to $AE$ and half of its length. Thus $MH$ is parallel to $GF$ and $MH = GF$. Hence $MGFH$ is a parallelogram. Then, $HQ=MH = GF$ and since $MH$ is parallel to $GF$, segment $HQ$ is also parallel to $GF$. Thus $HGFQ$ is a parallelogram and therefore $FQ$ is parallel to $HG$.
Segment $HG$ is a midpoint segment in triangle $MPQ$ and is therefore parallel to $QP$. Now, since $QF$ and $QP$ are both parallel to $HG$ and since there is only one line passing through $Q$ and is parallel to $HG$, lines $QF$ and $QP$ coincide and thus points $Q, F, P$ are collinear.