Problem
GEO-B3-M05-P013 Circumcircle as a Nine-Point Circle
Let \(I_a,I_b,I_c\) be the excenters of triangle \(ABC\). Prove that the circumcircle of \(ABC\) is the nine-point circle of triangle \(I_aI_bI_c\).
C. Hint 1. Recall that \(A,B,C\) are altitude feet of the excentral triangle.
D. Hint 2. Three altitude feet already determine the nine-point circle.
In the excentral triangle \(I_aI_bI_c\), the internal angle bisectors of the original triangle become altitudes. For instance, line \(AI\) is perpendicular to side \(I_bI_c\), so \(A\) is an altitude foot. Similarly, \(B\) and \(C\) are the other two altitude feet.
The nine-point circle of any triangle passes through its altitude feet. Hence the nine-point circle of \(I_aI_bI_c\) passes through \(A,B,C\). The unique circle through \(A,B,C\) is the circumcircle of the original triangle. Therefore the two circles coincide.
This connects excenters with the nine-point circle.