You can put this solution on YOUR website!
I presume you mean something like this, where the inner circle is the 9-point circle:
Suppose we extend HF to a point X on the circumcircle of ABC, and denote the midpoint of HC with Y (this must also lie on the 9-point circle, by definition):
By the Power of a Point theorem, HF*HY = k for some constant k. In addition, since HC = 2HY and HX = 2HF, then applying Power of a Point theorem again, HC*HX = 4k.
Intuitively, it would seem that the 9-point circle bisects each segment from the orthocenter to the circumcircle. This is because it bisects all of our known segments (such as HX, HA, etc.) and what we have found from Power of a Point theorem supports this. However I haven't yet found a way to prove this holds for all points since I chose arbitrary points on the circles and couldn't prove that two triangles were similar with ratio 1:2. Can you show this works for all points? You could try using similarity, or some other technique such as projective geometry or extending BE, AD to points Y and Z on the circumcircle, then connecting X, Y, and Z to create another triangle with orthocenter H. I'm sure there are several ways to accomplish this.