Six Point Circle
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.
What if applet does not run? 
Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander BogomolnyWhat if applet does not run? 
In ΔABC, the midpoints of the sides M_{A}, M_{B}, M_{C}, the feet H_{A}, H_{B}, H_{C} of the altitudes, and the midpoints A_{H}, B_{H}, C_{H} 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.
Let's, for example, consider the quadrilateral ABCH. The four midpoints of its sides are M_{A}, M_{C}, A_{H} and C_{H}. The four perpendiculars from the midpoints and their feet coalesce into two, H_{A} and H_{C}. (H_{A} is the foot of the perpendicular from A_{H} to BC and also of that from M_{A} to AH, and similarly for H_{C}.) Instead of eight concyclic points we only got six: M_{A}, M_{C}, A_{H}, C_{H}, H_{A} and H_{C}. But there are two additional quadrilaterals to take into account. Let's arrange the quadrilaterals and their 6 concyclic points in a table. In every case, I shall arrange the six points in two triples.

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 A_{H}M_{A}, B_{H}M_{B} and C_{H}M_{C}, 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 H_{A}H_{B}H_{C}. 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.
We may also remark that the 9 point circle in a triangle circumscribes its orthic (H_{A}H_{B}H_{C}), medial (M_{A}M_{B}M_{C}) and Euler's (A_{H}B_{H}C_{H}) triangles. Its radius equals one half of the circumradius.
The center of the 9point circle lies on Euler's line midway between the orthocenter and the circumcenter of ΔABC.
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.)
References
 H. S. M. Coxeter, Introduction to Geometry, John Wiley & Sons, 1961
 R. Honsberger, Mathematical Gems, II, MAA, 1976
Nine Point Circle
 Nine Point Circle: an Elementary Proof
 Feuerbach's Theorem
 Feuerbach's Theorem: a Proof
 Four 9Point Circles in a Quadrilateral
 Four Triangles, One Circle
 Hart Circle
 Incidence in Feuerbach's Theorem
 Six Point Circle
 Nine Point Circle
 6 to 9 Point Circle
 Six Concyclic Points II
 Bevan's Point and Theorem
 Another Property of the 9Point Circle
 Concurrence of Ten NinePoint Circles
 GarciaFeuerbach Collinearity
 Nine Point Center in Square
Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny63980834 