The Nine-Point Circle Theorem

A complete proof of the nine-point circle theorem. We show that nine special points (three side midpoints $M_A, M_B, M_C$, three orthocenter midpoints $E_A, E_B, E_C$, and three altitude feet $F_A, F_B, F_C$) all lie on a circle with center $N$ at the midpoint of $OH$ and radius $R/2$.

Prerequisites: orthocenter-proof, circumcenter-proof, h-reflect-midpoint-proof · Difficulty: advanced

The Theorem

For any triangle $ABC$, nine special points all lie on a single circle (the nine-point circle) with:

The nine points are:

Furthermore, each side midpoint and its corresponding orthocenter midpoint lie on opposite ends of a diameter (e.g., $M_A$ and $E_A$ are diametrically opposite).

Definitions

Circumcenter ($O$): The intersection of the perpendicular bisectors. Distance from $O$ to each vertex is $R$ (the circumradius).

Orthocenter ($H$): The intersection of the three altitudes.

Nine special points:

The Proof

Step 1: Triangle $ABC$

Let $N$ be the midpoint of $OH$. We will show that $N$ is equidistant from all nine points with distance $R/2$.

Step 2: The circumcenter $O$ and circumradius $R$

The circumcenter $O$ is the intersection of the perpendicular bisectors. It is equidistant from all three vertices at distance $R$, the circumradius.

Step 3: The orthocenter $H$

The orthocenter $H$ is where the three altitudes meet.

Step 4: The Euler line and midpoint $N$

$O$ and $H$ lie on the famous Euler line. Let $N$ be the midpoint of $OH$. We will prove $N$ is the center of a circle through six special points.

Step 5: Nine-Point Circle Theorem

For any triangle $ABC$, nine special points all lie on a single circle with center $N$ (the midpoint of $OH$) and radius $R/2$.

The nine points:

Furthermore, each side midpoint is diametrically opposite to its corresponding orthocenter midpoint.

Proof strategy. Apply the midpoint theorem twice to show $|NE_A| = |NM_A| = R/2$; symmetry extends this to all six midpoints. Then show $E_A M_A$ is a diameter and apply Thales' theorem to place the altitude feet on the circle.

Step 6: Part 1: Distance to $E_A$

Consider triangle $OHA$ with $N$ = midpoint of side $OH$ and $E_A$ = midpoint of side $AH$.

By the midpoint theorem, the segment connecting midpoints of two sides is parallel to the third side and half its length:

|NE_A| = \frac{1}{2}|OA| = \frac{R}{2}

Step 7: Part 2: $H'$ Lies on the Circumcircle

Reflection lemma (proved in the orthocenter chapter): reflecting the orthocenter $H$ over the midpoint of any side gives a point on the circumcircle, diametrically opposite to the opposite vertex.

Let $H'$ be the reflection of $H$ over $M_A$. By the lemma, $H'$ lies on the circumcircle opposite $A$, so $|OH'| = R$.

Step 8: Distance to $M_A$

In triangle $OHH'$, $N$ is the midpoint of $OH$ and $M_A$ is the midpoint of $HH'$ (by definition of reflection).

By the midpoint theorem:

|NM_A| = \frac{1}{2}|OH'| = \frac{R}{2}

We have now shown that both $E_A$ and $M_A$ lie on a circle centered at $N$ with radius $R/2$.

Step 9: By Symmetry: Six Points

By the same reasoning applied to vertices $B$ and $C$, the other four points ($M_B$, $M_C$, $E_B$, $E_C$) are also at distance $R/2$ from $N$. All six points lie on the circle.

Step 10: Diametrically Opposite Points

We showed in Part 1 that $NE_A$ is parallel to $OA$, and in Part 2 that $NM_A$ is parallel to $OH'$.

Since $H'$ is the point diametrically opposite $A$ on the circumcircle, $O$ lies on segment $AH'$ — so $OA$ and $OH'$ lie along the same line. Therefore $NE_A$ and $NM_A$ are both parallel to this line and both pass through $N$, so $M_A$, $N$, $E_A$ are collinear.

Combined with $NE_A = NM_A = R/2$, this makes $M_A E_A$ a diameter of the nine-point circle through its center $N$. The same holds for $(M_B, E_B)$ and $(M_C, E_C)$.

Step 11: Part 3: Distance to $F_A$

We now show the altitude feet also lie on the nine-point circle using the diametrically opposite property.

Since $E_A M_A$ is a diameter, by Thales' theorem, any point $P$ on the circle satisfies $\angle E_A P M_A = 90°$. Conversely, if $\angle E_A P M_A = 90°$, then $P$ lies on the circle.

Claim: $\angle E_A F_A M_A = 90°$

The altitude from $A$ is perpendicular to $BC$. Both $E_A$ (midpoint of $AH$) and $F_A$ (altitude foot) lie on this altitude. Since $M_A$ is on $BC$, we have $E_A F_A \perp F_A M_A$, so $\angle E_A F_A M_A = 90°$.

By Thales' theorem (converse), $F_A$ lies on the nine-point circle.

Step 12: The Nine-Point Circle: Complete

By the same argument applied to vertices $B$ and $C$:

We have proven that all nine special points lie on a circle with center $N$ (midpoint of $OH$) and radius $R/2$: three side midpoints, three orthocenter midpoints, and three altitude feet.

Conclusion

We have proven geometrically that all nine special points lie on a circle with:

The key tools were:

Historical Note

The Euler line was discovered by Leonhard Euler in 1765. The nine-point circle was discovered independently by Karl Wilhelm Feuerbach, Charles Brianchon, and Jean-Victor Poncelet in the early 19th century. Feuerbach also proved the remarkable theorem that the nine-point circle is tangent to the incircle and all three excircles.