Nine-Point Circle and Center (Midpoint of OH)

One of the more interesting circles every triangle possesses is the nine-point circle, a circle whose proof of existence combines many different geometric configurations into one diagram. The nine point circle for an arbitrary triangle is given below.

Okay, well that's great and all... we have a circle, but where are the nine points? This post here will delve into two key lemmas that show exactly why this circle is called the nine-point circle.

Before you explore this section, I highly suggest becoming familiar with cyclic quadrilaterals. The relevant expii topic can be found here.

Awesome!

The nine-point circle makes use of two of the four special points in a triangle: the orthocenter and the circumcenter. They are designated as $H$ and $O$ in the diagram below.

Great! Now that you know that, let's get to our first key lemma: The reflections of the orthocenter $H$ across the sides $AB$, $BC$, $CA$ of $\triangle ABC$ all lie on the circumcircle. A diagram (simplified of course) is shown below.

Why must this be true? Think about it for a little bit before continuing.

The proof is simple by angle chasing. Let $K$ be the image of $H$ after reflection as shown. Note that $\angle KBD=\angle HBD$ and $\angle KCD=\angle HCD$ by the nature of reflection, so $\triangle HBC=\triangle KBC$. Furthermore, since $AEHF$ is a cyclic quadrilateral we have $\angle EHF=\pi-\angle A$. Therefore $$\angle BKC=\angle BHC=\angle EHF=\pi-\angle A,$$ establishing the desired cycliclity. The proof can be extended to each of the other sides of $\triangle ABC$ as desired.

Confused? Read the proof above until you're fully comfortable with it. Once you are, press "Continue".

The next lemma follows as almost a corollary. Suppose we reflect $H$ not over the side $BC$ but instead over its midpoint $M$. Then the reflection point still lies on the circumcircle!

And in fact, the proof is almost exactly the same as the proof for the previous case!

Let $L$ be the image under reflection across the midpoint $M$. By the definition of reflection, $HM=ML$ and $\angle HMB=\angle CML$. Combining this with $BM=MC$ gives $\triangle HMB=\triangle LMC\implies \angle BHL=\angle HLC$. Similar reasoning gives $\angle CHM=\angle HMB$, and adding both equalities together gives $\angle BHC=\angle BLC$. Now use the final part of the previous proof to show that $L$ lies on the circumcircle.

(Food for thought: although it is not important to the proof of the nine-point circle's existance, it turns out that $AL$ is a diameter of the circumcircle! Can you explain this?)

Are you ready to move on? We're now going to see how all of this has to do with that mysterious circle.

Okay, now we're ready to tie everything together, and to do this, we're going to need a geometric technique called homothety (or dilation). The name seems pretty fancy, but in reality it just means we're going to scale stuff. An explanation can be found here.

Consider the homothety centered at $H$ with scale factor $\frac12$. What happens to various points on the circumcircle under this dilation? A somewhat-boring observation is that $A$ is taken to the midpoint of $\overline{HA}$. More interesting is what happens to $K$ and $L$. Since $D$ (the foot of the altitude from $A$) is the midpoint of $\overline{HK}$ and $M$ (the midpoint of $\overline{BC}$) is the midpoint of $\overline{HL}$, $K$ and $L$ are actually taken to $H$ and $M$ respectively! And since homothety transforms circles to circles, these three points all still lie on corresponding positions of a (new) circle!

Applying this cyclically, we get the essence of the nine-point circle: The midpoints of the sides, the feet of the altitudes, and the midpoints of $\overline{HA}$, $\overline{HB}$, $\overline{HC}$ all lie on a single circle known as the nine-point circle of $\triangle ABC$.

Now that we know what the nine-point circle is, we will go on and explore two interesting properties it has.

First, the nine-point center $N$ (i.e. the center of the nine-point circle) is the midpoint of $OH$. (This is actually not hard to prove; just consider the same homothety as before.) This implies that $N$ lies on the Euler Line of $\triangle ABC$.

A second, arguably awesome fact is that the nine-point circle is tangent to the incircle of any triangle. The point at which the nine-point circle and incircle are tangent to each other is called the Feuerbach Point and the common tangent is called the Feuerbach Line (shown in dashed blue in the diagram).

And that's it! There are many other properties and explorations that the nine-point circle can take you; this website is a great starting point.

Happy nine-point circle hunting!