A triangle and its altitudes define three quadrilaterals with orthogonal diagonals. Strangely, the 8 point circles of the three quadrilaterals coincide. Thus we get a circle in the triangle with 9 points on it. See for yourself.
In ABC, the midpoints of the sides MA, MB, MC, the feet HA, HB, HC of the altitudes, and the midpoints AH, BH, CH of the segments connecting the orthocenter with the vertices, lie on a circle, known as the 9 point circle.
As was mentioned at the outset, a triangle ABC and its altitudes define three quadrilaterals with orthogonal diagonals: ABCH, BCAH, CABH. In each of the quadrilaterals, we may identify the 8 points that we know are concyclic, i.e. lie on a circle.
Each of the triples alone defines a circle. Since the triples come in pairs that comprise six concyclic points, each pair of the triples define the same circle, which exactly means that all three triples - 9 points in all - all lie on the same circle.
The center of the common circle lies at the intersection of the segments AHMA, BHMB and CHMC, which are thus concurrent. Each serves as a diameter of the 9 point circle.
The nine point circle is the circumcirlce of the orthic triangle HAHBHC. The four points A, B, C and H taken three at a time form four triangles - ABC, ABH, BCH, CAH - with the property that the remaining point serves as the orthocenter of the corresponding triangle. Any four points with this property form an orthocentric system. The four triangles share the same orthic triangle and the same nine point circle.
The incenter of a triangle together with its three excenters form the configuration of four points that fits the foregoing discussion. In particular, the excenters and the incenter of a triangle form an orthocentric system.
(An even easier proof of the existence of the nine point circle could be found elsewhere at the site.)