Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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 at 19:36
    
Thanks for your comment, I'll try to add it soon. Thanks –  akusaja Mar 26 at 19:49
    
Just check: according to your drawing, the circles are $\;AHM\,,\,CHN\;$ ... –  DonAntonio Mar 26 at 20:16
    
@DonAntonio, thanks for the correction. Here's the new image: i62.tinypic.com/2cmsvty.png –  akusaja Mar 27 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 at 9:58
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.