Olympiad Geometry Problem #46: Circumcenter, Parallel, Orthocenter

Here is a very beautiful problem with a simple starting configuration, that I found on the Art of Problem Solving forum. Please let me know if you find a simpler solution! Link below.

[helo]

how would you prove AH=2*distance from O to BC?

[ManojSingh-wcxp]

Maybe drop a perpendicular PM from P on to BC and then simply KPFM is cylci suffices.

[dicksonchang]

<KFH=<KHF=<ABD
KF/PB=KH/PB=the distant from O to BC /PB=sin(<PBC)=sin(<FBC)=GF/GB which implies that triangle KFG is similar to triangle PBC (AA)
by easy arguement we can show triangle KPG is similar to triangle FBG (SAS)
which proves that <PKC=<BFH=90°