A homothety with center \(A\) with ratio \(\frac{1}{2}\) maps \(BC\) to the midline \(B'C'\). This maps \(M\) to the circumcenter of \(H'B'C'\) where \(H'\) is the midpoint of \(AH\). But it is well-known that the circumcenter we are looking for is the nine-point center, so done.

This is also very easy with complex numbers; since M, N, P are $b+c, c+a, a+b$, respectively, and $h=a+b+c$, we can easily find that the intersection of all of these lines is $\frac{a+b+c}{2}$.

$AHMO$, etc. are parallelograms, thus we are done. Best regards, sunken rock

It's very easy problem. Show that R(ABC)=R(AHC) and CH=PO ==> CP passing trough mindpoint of OH.

center of $\odot (BHC)$ is the reflection of $O$ over $BC$, which forms $AHMO$ (parallelogram) $\implies$ $AM,BN,CP$ concur at $N_9$ which is the nine point circle WRT $\Delta ABC$ $\in OH$

By angle chasing, we can quickly see that PNM is similar to CBA and by considering the circles, we can quickly get angle HMN=HBA=HAC=HNM so HN=HM, so PMN is congruent to CAB as well and then we basically have H is the circumcentre of PMN and O is the orthocentre of PMN then OM=AH, HM=AO, so AHMO is a parallelogram and AM intersects OH at a point T where T is the mid pt of OH and by symmetry all other lines meet at T as well so the four of them are concurrent