Let ABC be an acute triangle with altitudes AD and BE. The
intersection of AD and BE is H. Points F and G make CADF
and CBEG into parallelograms. M is the midpoint of FG.
Ray MC intersects the circumcircle of ΔABC again at point N.

Prove that ANBH is a parallelogram.
  

There's a lot going on in this problem, but I feel the key may be to add a line or two.  Where?  It's easy to prove angles ANB and BNA are equal but that's all I've got.

Posted by Jer on 2017-09-11 20:52:59