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Given: triangle ABC is acute triangle.

M and N are midpoints of AB and BC respectively, while BH is altitude of triangle ABC.

Circles AHN and CHM meet at point P. (P is not same with H)

How to prove that PH is containing midpoint of side MN?

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as you've been asking several questions in geometry lately it'd be a good idea, imo, if you find some way to add a diagram, otherwise it can be hard to follow the description and maybe not many people will take the work to read and decypher the question... –  DonAntonio Mar 26 '14 at 19:36
Thanks for your comment, I'll try to add it soon. Thanks –  akusaja Mar 26 '14 at 19:49
Just check: according to your drawing, the circles are $\;AHM\,,\,CHN\;$ ... –  DonAntonio Mar 26 '14 at 20:16
@DonAntonio, thanks for the correction. Here's the new image: i62.tinypic.com/2cmsvty.png –  akusaja Mar 27 '14 at 9:21
@akusaja: I had thought the fix would be to switch the places of $M$ and $N$; proof of the result is pretty straightforward for that case. (You can show that $MN$ is tangent to both circles.) However, according to my GeoGebra sketch, it appears that the result also holds when leaving $M$ and $N$ where they are; I don't have a proof of that case (yet). I'm afraid this brings up a familiar question: What does the original question actually say? –  Blue Mar 27 '14 at 9:58

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