Nine-point Circle

The nine-point circle of a triangle is a circle going through 9 key points:

• the three midpoints of the sides of the triangle (blue in the below picture),
• the three feet of the altitudes of the triangle (yellow in the below picture), and
• the three midpoints from the vertices to the orthocenter of the triangle (green in the below picture).

The nine-point circle satisfies several important and interesting properties, besides the surprising fact that it exists at all.

Let the midpoint of $AB$ be $M_c$, the midpoint of $BC$ be $M_a$, and the midpoint of $CA$ be $M_b$. Similarly, let the foot of the altitude from $A$ be $A_h$, the foot of the altitude from $B$ be $B_h$, and the foot of the altitude from $C$ be $C_h$. Let the intersection of $A_h$, $B_h$, and $C_h$ be H. Finally, let the midpoint of $AH$ be $D$, the midpoint of $BH$ be $E$, and the midpoint of $CH$ be $F$. The goal is to show that $M_a, M_b, M_c, A_h, B_h, C_h, D, E$ and $F$ all lie on a circle.

Firstly, since $M_b$ and $M_c$ are midpoints of $AC$ and $AB,$ respectively, the segment $M_bM_c$ is parallel to the segment $BC$. In addition, $E$ and $F$ are midpoints of $BH$ and $CH,$ respectively, so the segment $EF$ is parallel to the segment $BC$ as well. Therefore, $M_bM_c$ and $EF$ are parallel.

Similarly, since $M_c$ and $E$ are midpoints of $AB$ and $BH,$ respectively, the segment $M_cE$ is parallel to the segment $AH$. Analogously, $M_bF$ is parallel to $AH$, so $M_cE$ and $M_bF$ are parallel as well.

Hence $M_cM_bFE$ is a parallelogram. But $M_cE$ is parallel to $AH$, which is the same line as $AA_h$. Furthermore, $AA_h$ is perpendicular to $BC$, and since $M_cM_b$ is parallel to $BC$, $AA_h$ is perpendicular to $M_cM_b$ as well. Therefore, $M_cE$ is perpendicular to $M_cM_b$, meaning that $M_cM_bFE$ is a rectangle.

By similar logic, $M_bM_aED$ and $M_aM_cDF$ are rectangles as well, so $M_c, D, M_b, F, M_a$, and $E$ all lie on the same circle.

Let $M_cF$ and $DM_a$ intersect at a point $N$, which is the center of the circle going through the above six points $($since $M_cDFM_a$ is a rectangle$).$ Therefore, $N$ is the midpoint of $DM_a$. But $\angle DA_hM_a=90^{\circ}$, so $N$ is the circumcenter of triangle $DA_hM_a$. Thus $A_h$ lies on the same circle as the above six points do, and by identical logic so do $B_h$ and $C_h$.

Therefore, the points $M_a, M_b, M_c, A_h, B_h, C_h, D, E,$ and $F$ all lie on a circle, as desired.

The center of the nine-point circle, known as the nine-point center $N$, lies on the Euler line of the triangle, and is the midpoint of the section of the Euler line from the orthocenter $H$ to the circumcenter $O$.

The nine-point center $N$ is the midpoint of segment $OH$. Also shown: the centroid $G$

The nine-point circle is also the circumcircle of the orthic triangle and the medial triangle (the triangle whose vertices are the three midpoints).

Furthermore, due to symmetry, we have the following:

The nine-point circles of triangles $ABC, ABH, BCH,$ and $CAH$ are all the same.

As a consequence, the circumradii of those four triangles are equal, and as a corollary, we have the following:

$NA^2+NB^2+NC^2+NH^2=3R^2$, where $R$ is the circumradius of the triangle.

Similarly, for any point $P$ on the nine-point circle,

$$PA^2+PB^2+PC^2+PH^2=4R^2.$$

Consequently, the radius of the nine-point circle is exactly half the circumradius of the triangle:

$$r_N = \frac{1}{2}R.$$

Remarkably, if $P$ is a point for which $A, B, C$, and $P$ do not form an orthocentric system, then the nine-point circles of $\triangle ABC, \triangle ABP, \triangle BCP$, and $\triangle CAP$ all concur at a single point.

Similarly, the nine-point circle is externally tangent to each of the three excircles, and internally tangent to the triangle's incircle. The point at which the nine-point circle is tangent to the incircle is called the Feuerbach point.

• Euler's Line
• Incircles and Excircles