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.

Proof of Existence

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.

Properties

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 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.

Contents

Proof of Existence

Properties

See Also