6 to 9 Point Circle: What Is This About?
A Mathematical Droodle

1 May 2015, Created with GeoGebra


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Copyright © 1996-2017 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

  • Ma, Mb, Mc - the midpoints of sides BC, CA, and AB,
  • Ha, Hb, Hc - the feet of altitudes to sides BC, CA, and AB,
  • Ah, Bh, Ch - the midpoints of AH, BH, CH - the so called Euler points.

six point circle to nine point circle

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 (Oa), (Ob), (Oc), which are the circumcircles of triangles HHaMa, HHbMb, HHcMc. Of course the three centers Oa, Ob, Oc are the midpoints of HMa, HMb, HMc. Two circles (Ob), (Oc) share one common point at H. Their other common point is the reflection of H in ObOc therefore it is on altitude AHa. (This is because ObOc ⊥ HHa.) Similarly we can show: three circles (Oa), (Ob), (Oc) share one common point at H and another common points are on the Cevians (altitudes) AHa, BHb, CHc. By the six concyclic points theorem, Ha, Hb, Hc, Ma, Mb, Mc 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., Ch, Ah, Mb and, of course, through the feet of the altitudes, which happen to be Ha, Hb, Hc. The circles (N) formed for triangles ABH also BCH, also pass through Ha, Hb, Hc so that all three coincide. The latter two pass through Ah, Bh, Mc and Bh, Ch, Ma, respectively. It follows that the common (N) circle passes through all the nine points.

Nine Point Circle

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Copyright © 1996-2017 Alexander Bogomolny


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