Nine point circle

Triangle ABC with the nine point circle in light orange

The nine point circle (also known as Euler's circle or Feuerbach's circle) of a given triangle is a circle which passes through 9 "significant" points:

That such a circle exists is a non-trivial theorem of Euclidean geometry.

The center of the nine point circle is the nine-point center and is usually denoted $N$.

The nine point circle is tangent to the incircle, has a radius equal to half the circumradius, and its center is the midpoint of the segment connecting the orthocenter and the circumcenter, upon which the centroid also falls.

It's also denoted Kimberling center $X_5$.


First Proof of Existence

Since $O_c$ is the midpoint of $AB$ and $E_b$ is the midpoint of $BH$, $O_cE_b$ is parallel to $AH$. Using similar logic, we see that $O_bE_c$ is also parallel to $AH$. Since $E_b$ is the midpoint of $HB$ and $E_c$ is the midpoint of $HC$, $E_bE_c$ is parallel to $BC$, which is perpendicular to $AH$. Similar logic gives us that $O_bO_c$ is perpendicular to $AH$ as well. Therefore $O_bO_cE_bE_c$ is a rectangle, which is a cyclic figure. The diagonals $O_bE_b$ and $O_cE_c$ are diagonals of the circumcircle. Similar logic to the above gives us that $O_aO_cE_aE_c$ is a rectangle with a common diagonal to $O_bO_cE_bE_c$. Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle $O_aO_bE_aE_b$ is also on the circle.

We now have a circle with the points $O_a$, $O_b$, $O_c$, $E_a$, $E_b$, and $E_c$ on it, with diameters $O_aE_A$, $O_bE_b$, and $O_cE_c$. We now note that $\angle E_aH_aO_a=\angle E_bH_bO_b=\angle E_cH_cO_c=90^{\circ}$. Therefore $H_a$, $H_b$, and $H_c$ are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore the nine points are on the circle, and the nine-point circle exists.

Second Proof of Existence

We know that the reflection of the orthocenter about the sides and about the midpoints of the triangle's sides lie on the circumcircle. Thus, consider the homothety centered at $H$ with ratio ${1}/{2}$. It maps the circumcircle of $\triangle ABC$ to the nine point circle, and the vertices of the triangle to its Euler points. Hence proved.

See also


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