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?|
|What if applet does not run?|
|In ΔABC, 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.
Let's, for example, consider the quadrilateral ABCH. The four midpoints of its sides are MA, MC, AH and CH. The four perpendiculars from the midpoints and their feet coalesce into two, HA and HC. (HA is the foot of the perpendicular from AH to BC and also of that from MA to AH, and similarly for HC.) Instead of eight concyclic points we only got six: MA, MC, AH, CH, HA and HC. 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 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.
We may also remark that the 9 point circle in a triangle circumscribes its orthic (HAHBHC), medial (MAMBMC) and Euler's (AHBHCH) triangles. Its radius equals one half of the circumradius.
The center of the 9-point 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.)
- 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 9-Point 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 9-Point Circle
- Concurrence of Ten Nine-Point Circles
- Garcia-Feuerbach Collinearity
- Nine Point Center in Square