A triangle $\Delta ABC$ has its orthocenter $H$ distinct from its vertices and from the circumcenter $O$. Denote $M$, $N$, $P$ the circumcenters of the triangles $\Delta HBC$, $\Delta HCA$, and $\Delta HAB$ repectively.
Prove that the lines $AM$, $BN$, $CP$ and $OH$ are concurrent.
I draw the figure but unable to proceed. Thanks in advance.
1) $R_{ABC}=R_{HBC}=R_{AHC}=R_{ABH}$
Then $$PB=BM=MC=CN=NA=AP$$
2) Let $K$ are midpoint of $HO$
Then $PC \cap BN \cap AM = K$
We'll prove that the intersection point is the center of the nine-point circle, $T$. Let $M_1$, $M_2$ be the midpoints of $AC$ and $AB$. Additionally let $AC$ intersect the $\odot AHB$ for a second time at $L$. Then we have that:
$$\angle CLB = \angle ALB = \pi - \angle AHB = \pi - \angle ABC - \angle BAC = \angle BCL$$
Therefore the if $E$ is the foot of the altitute from $B$ to $AC$ we have that $CL = 2CE$, combined with the obvious $CA = CM_1$ and $CB = CM_2$ we have that there is a homothety sending $\odot AHBL$ to the nine-point circle. Therefore $C$ lies on the line connecting the center of those two circles, i.e. $P,T,C$ are collinear. Similarly $M,T,A$ and $N,T,B$ are collinear. Additionally it's well-known that $T$ lies on the line $OH$