Fermat Problem

Theorem
Let $\triangle ABC$ be a triangle

Let the vertices of $\triangle ABC$ all have angles less than $120 \degrees$.

Let $\triangle ABG$, $\triangle BCE$ and $\triangle ACF$ be equilateral triangles constructed on the sides of $ABC$.

Let $AE$, $BF$ and $CG$ be constructed.

Let $P$ be the point at which $AE$, $BF$ and $CG$ meet.


 * FermatPointConstruction.png

Then $P$ is the Fermat point of $\triangle ABC$.

If one of vertices of $\triangle ABC$ be of $120 \degrees$ or more, then that vertex is itself the Fermat point of $\triangle ABC$.