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Parallelograms (Posted on 2017-09-09) Difficulty: 4 of 5

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.

See The Solution Submitted by Bractals    
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Hints/Tips re: May I have a hint, please? | Comment 2 of 5 |
(In reply to May I have a hint, please? by Jer)

Do you have a program like "Geometer's Sketchpad"
There is no way I could have cracked this problem without it.

If you made a nice sketch you will notice that the ray MC passes through the point O (circumcenter of triangle ABC). I thought this was how I constructed my sketch. But no, the way point M is constructed it is forced to lie on line CO. After seeing this I was hoping to drop the difficulty to 3 or 2.

I have started to put together my solution. Should have it out in a couple of days.

Hope the above helps.

  Posted by Bractals on 2017-09-12 11:39:34
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