Nine-point circle

Dependencies: Triangle midsegment theorem (the midsegment $\parallel$ third side, length $=$ half the base), Thales converse (right angle $\Rightarrow$ on the circle with the hypotenuse as diameter), Parallelogram tests (one pair of opposite sides parallel and equal $\Rightarrow$ parallelogram; rectangle $=$ parallelogram $+$ one right angle).

Statement

Let $\triangle ABC$ be a non-degenerate triangle with orthocenter $H$. Consider the following nine points:

• the midpoints $M_A, M_B, M_C$ of the three sides (with $M_A$ the midpoint of $BC$, and so on),
• the feet $H_A, H_B, H_C$ of the three altitudes (with $H_A$ the foot of the altitude from $A$ on $BC$, and so on),
• the midpoints $N_A, N_B, N_C$ of the segments joining the orthocenter $H$ to each of the three vertices $A, B, C$ (with $N_A$ the midpoint of $\overline{AH}$, and so on).

Then all nine points lie on a single circle — called the nine-point circle of $\triangle ABC$.

The center $N$ of the nine-point circle is the midpoint of the circumcenter $O$ and the orthocenter $H$ (it lies on the Euler line: O, G, H collinear), and its radius is exactly half the circumradius, $\tfrac12 R$.