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 Parallelograms (Posted on 2017-09-09)

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 No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: May I have a hint, please? | Comment 3 of 5 |

"It's easy to prove angles ANB and BNA are equal..."

Unless you are using directed angles, ANB and BNA are always equal.

I'm still having a problem showing that the ray MC passes through the circumcenter of triangle ABC.

If I could do that, then proving that ANBH is a parallelogram is easy.

If I could prove ANBH is a parallelogram, then proving that the
ray MC passes through the circumcenter of triangle ABC is easy.

 Posted by Bractals on 2017-09-18 22:38:15

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