# 6 to 9 Point Circle: What Is This About?

A Mathematical Droodle

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Copyright © 1996-2018 Alexander Bogomolny

The existence of the 9-point circle for a given triangle, may be established in a variety of ways. In ΔABC, the 9-point circles passes through

- M
_{a}, M_{b}, M_{c}- the midpoints of sides BC, CA, and AB, - H
_{a}, H_{b}, H_{c}- the feet of altitudes to sides BC, CA, and AB, - A
_{h}, B_{h}, C_{h}- the midpoints of AH, BH, CH - the so called*Euler points*.

Bui Quang Tuan came up with a novel proof based on the recent theorem of six concyclic points. Let, as usual, H denote the orthocenter of ΔABC. The proof begins with a construction of three circles (O_{a}), (O_{b}), (O_{c}), which are the circumcircles of triangles HH_{a}M_{a}, HH_{b}M_{b}, HH_{c}M_{c}. Of course the three centers O_{a}, O_{b}, O_{c} are the midpoints of HM_{a}, HM_{b}, HM_{c}. Two circles (O_{b}), (O_{c}) share one common point at H. Their other common point is the reflection of H in O_{b}O_{c} therefore it is on altitude AH_{a}. (This is because _{b}O_{c} ⊥ HH_{a}.)_{a}), (O_{b}), (O_{c}) share one common point at H and another common points are on the Cevians (altitudes) AH_{a}, BH_{b}, CH_{c}. By the six concyclic points theorem, H_{a}, H_{b}, H_{c}, M_{a}, M_{b}, M_{c} are concyclic on a circle, say, (N). In words,

In a given triangle, the feet of altitudes and the midpoints of the sides are concyclic.

Applying this statement to, say, ΔCAH we see that its circle (N) passes through the midpoints of its sides, viz., C_{h}, A_{h}, M_{b} and, of course, through the feet of the altitudes, which happen to be H_{a}, H_{b}, H_{c}. The circles (N) formed for triangles ABH also BCH, also pass through H_{a}, H_{b}, H_{c} so that all three coincide. The latter two pass through A_{h}, B_{h}, M_{c} and B_{h}, C_{h}, M_{a}, respectively. It follows that the common (N) circle passes through all the nine points.

### Nine Point Circle

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Copyright © 1996-2018 Alexander Bogomolny