### Motivation

Someone on Discord asked about the existence of the nine-point circle. It’s well-known that that can be proved by homothety.

### Little reminder about homothety

Homothety preserves angles (and thus parallel lines). Homothetic polygons are similar, so the ratio of the corresponding sides is the same. Considering the radii of a circle under a homothety, we see that a homothety maps a circle to another circle.

### Notation

*H*: orthocenter*G*: centroid*O*: circumcenter*ω*: circumcircle*H*: feet of altitude with respect to_{A}*A*.*M*: midpoint of side_{A}*a*.*E*: Euler point with respect to_{A}*A*. (i.e. midpoint of*A*and*H*)

### Problem

The second proof for nine-point circle on AoPS starts with a
proved fact that the reflection of *H* about *a* and *M _{A}* lie on

*ω*.

The first part is easy. I changed the definition of *H _{A}*’ from
“reflect(

*H*,

*a*)” to “

*AH*∩

*ω*”.

In either definition, we can prove △*BHC* ≅ △*BH _{A}C*. In my
definition, use ∠

*CBH*’ = ∠

_{A}*CAH*= ∠

*CBH*= 90° −

*C*.

I found the second part is harder because I used *A*’ := *HM _{A}* ∩

*ω*" instead of “reflect(

*H*,

*M*)”. After thinking about that for hours, I gave up and searched for “nine point circle homothety” for answer, but the information was too advanced for me. Finally, I searched with more relevant keywords “orthocenter reflection”, and I found the hints from this relevant Math.SE question.

_{A}### Solution 1

In the linked solution on Math.SE, *A*’ := reflect(*H*, *M _{A}*). It’s
easy to see that

*BA*’

*CH*is a parallelogram, so ∠

*BA*’

*C*= ∠

*BHC*=

*B*+

*C*= 180° −

*A*. As a result,

*A*’ ∈

*ω*.

### Solution 2

Using my definition of *A*’, just reverse the reasoning above.

We see that *A*’*C* ⟂ *AC* and *A*, *A*’, *C* ∈ *ω*, so *AA*’ is a diameter for
*ω*. This shows that *AH* = 2*OM _{A}*.

### Solution 3

On this post on AoPS, *A*’ := reflect(*A*, *O*).

### Corollary

The nine-point circle is the image of the circumcircle *ω* under the homothety
h(*H*, 1/2).