Let $ABC$ be a acute triangle and three altitude foot from $A,B,C$ are $D,E,F$ respectively. $H$ is the orthorcenter of triangle $ABC$. O is center of circumcircle of triangle $BHC$. Let $N=DF\cap BH, M=DE\cap CH$. Prove that $AO\perp MN$.
Here are what i did:
- Let $K=EF\cap MN$ and $L=AD\cap EF$, we have $(K,L,F,E)=-1$. Let $K'=EF\cap BC$ then $(K',L,E,F)=1$. Therefore $K\equiv K'$.
- O is a symmetrical to the center of $(ABC)$ with respect to the straight line $BC$.
I'm stuck here. Somebody can help me!
Hint: As shown in figure KC is radical axis of two circles. KA=KG are tangent on two circles(property of radical axis). You have to show that intersection of circles centers at Q and K intersect on MN that would mean $MN\bot AO$.