For an arbitrary triangle $ABC$, mark these nine points:

• the midpoints of the three sides;
• the feet of the three altitudes;
• the midpoints between each vertex and the orthocenter $H$.

Prove that all nine points lie on one circle.

Hints

Open one at a time

Start with the medial triangle.

Look for cyclic quadrilaterals created by right angles.

Solution

Best opened after a real attempt

Begin with six points

Let $D,E,F$ be the altitude feet and $L,M,N$ the side midpoints. Since $\mathrm{\angle }BDC=\mathrm{\angle }BEC={90}^{\circ }$, points $B,C,D,E$ are concyclic. Similar right-angle arguments connect each altitude foot to the medial triangle.

A compact route is to apply a homothety centered at $H$ with scale factor $1/2$. It sends $A,B,C$ to the three midpoints between the vertices and $H$, and sends the circumcircle of $ABC$ to a circle of half the circumradius.

The center of that image circle is the midpoint of $OH$, where $O$ is the circumcenter. One then verifies that the side midpoints and altitude feet are at the same distance $R/2$ from this center.

All nine points therefore lie on the circle centered at the midpoint of $OH$ with radius $R/2$.