The Nine-Point Circle Theorem

A complete proof of the nine-point circle theorem. We show that nine special points — the three side midpoints, the three midpoints of the segments joining each vertex to the orthocenter, and the three altitude feet — all lie on one circle, with center N at the midpoint of OH and radius R/2.

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

For any triangle $ABC$, nine special points all lie on a single circle (the nine-point circle): the three side midpoints, the three midpoints of the segments joining each vertex to the orthocenter, and the three altitude feet. Its center is $N=\frac{1}{2}(O+H)$ and its radius is $\frac{R}{2}$.

Strategy: apply the midpoint theorem twice to show $|N{E}_A|=|N{M}_A|=R/2$; 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 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:

• Three side midpoints $M_A,M_B,M_C$ on $BC$, $CA$, $AB$
• Three orthocenter midpoints $E_A,E_B,E_C$, the midpoints of $AH$, $BH$, $CH$
• Three altitude feet $F_A,F_B,F_C$

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

Proof strategy. Apply the midpoint theorem twice to show $|N{E}_A|=|N{M}_A|=R/2$; 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:

Step 7: Part 2: $H^{\prime }$ 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 $H^{\prime }$ be the reflection of $H$ over $M_A$. By the lemma, $H^{\prime }$ lies on the circumcircle opposite $A$, so $|O{H}^{\prime }|=R$.

Step 8: Distance to $M_A$

In triangle $OH{H}^{\prime }$, $N$ is the midpoint of $OH$ and $M_A$ is the midpoint of $H{H}^{\prime }$ (by definition of reflection).

By the midpoint theorem:

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 $N{E}_A$ is parallel to $OA$, and in Part 2 that $N{M}_A$ is parallel to $O{H}^{\prime }$.

Since $H^{\prime }$ is the point diametrically opposite $A$ on the circumcircle, $O$ lies on segment $A{H}^{\prime }$ — so $OA$ and $O{H}^{\prime }$ lie along the same line. Therefore $N{E}_A$ and $N{M}_A$ are both parallel to this line and both pass through $N$, so $M_A$, $N$, $E_A$ are collinear.

Combined with $N{E}_A=N{M}_A=R/2$, 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 $C$:

• $\angle E_B{F}_B{M}_B=90°$, so $F_B$ is on the nine-point circle
• $\angle E_C{F}_C{M}_C=90°$, so $F_C$ is on the nine-point circle

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.

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

• Center: $N=\frac{1}{2}(O+H)$, the midpoint of the Euler line segment
• Radius: $\frac{R}{2}$, half the circumradius

The key tools were:

• Midpoint theorem applied to triangles $OHA$ and $OH{H}^{\prime }$: proving $|N{E}_A|=|N{M}_A|=R/2$
• Reflection lemma (proved in the orthocenter chapter): the reflection $H^{\prime }$ of $H$ over $M_A$ lies on the circumcircle, so $|O{H}^{\prime }|=R$
• Thales' theorem: proving the altitude feet lie on the circle using the diameter property

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. ∎

Notes

Definitions

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

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

Nine special points:

• $M_A,M_B,M_C$: midpoints of the sides $BC$, $CA$, $AB$
• $E_A,E_B,E_C$: midpoints of segments $AH$, $BH$, $CH$
• $F_A,F_B,F_C$: feet of the altitudes from $A$, $B$, $C$

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).