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 :)