Relating the incenters of the original and medial triangles.

Question

Let I be the incenter of △ABC. If I is also the incenter of the medial triangle of △ABC, show that △ABC must be equilateral.

I'm thinking a place to start would be to show the distance between AC and WU is the same as the distance between AB and VU, but for one, I'm not sure if this is a good starting point, and two, I can't figure out how to do it in the first place.

Any help would be appreciated.

Answer 1

Here is a solution using Barycentric Coordinates.

First note that the incentre of $\triangle{DEF}$ is the spieker centre $S_p$ of $\triangle{ABC}$. With $\triangle{ABC}$ is the reference triangle we have,

$$S_p\equiv{}I\implies{}\dfrac{1}{4s}\left(b+c,c+a,a+b\right)\equiv{}\dfrac{1}{2s}\left(a,b,c\right).$$

Therefore $2a=b+c,2b=c+a$ and $2c=a+b$. Hence the result follows.

Reference: Barycentric Coordinates