# Nine-point circle

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:

- The three feet of the altitudes of the triangle.
- The three midpoints of the edges of the triangle.
- The three midpoints of the segments joining the vertices of the triangle to its orthocenter. (These points are sometimes known as the Euler points of the triangle.)

"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." -hankinjg

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 .

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 .

## First Proof of Existence

Since is the midpoint of and is the midpoint of , is parallel to . Using similar logic, we see that is also parallel to . Since is the midpoint of and is the midpoint of , is parallel to , which is perpendicular to . Similar logic gives us that is perpendicular to as well. Therefore is a rectangle, which is a cyclic figure. The diagonals and are diagonals of the circumcircle. Similar logic to the above gives us that is a rectangle with a common diagonal to . Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle is also on the circle.

We now have a circle with the points , , , , , and on it, with diameters , , and . We now note that . Therefore , , and 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 with ratio . It maps the circumcircle of to the nine-point circle, and the vertices of the triangle to its Euler points. Hence proved.

## Common Euler circle

Let an acute-angled triangle with orthocenter be given.

be the point on opposite

Points and such that is a parallelogram. The line intersects at the points and

Prove that triangles and has common Euler (nine-point) circle.

* Proof*
Denote is midpoint

Let’s consider Circumcenter of point is the midpoint point is the midpoint

Denote the centroid of

is the centroid of

Denote the midpoint of is the midpoint of

is the centroid of

Point is the circumcenter of is the orthocenter of

The triangles and has common circumcircle and common center of Euler circle (the midpoint of ) therefore these triangles has the common Euler circle.

**vladimir.shelomovskii@gmail.com, vvsss**

## See also

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