Fermat$-$Torricelli theorem

Theorem (Fermat$\mathrm{-}$Torricelli).  Let all angles of a triangle $ABC$ be at most ${120}^{\circ }$.  Then the inner point $F$ of the triangle which makes the sum $AF+BF+CF$ as little as possible, is the point from which the angle of view of every side is ${120}^{\circ }$.

Proof.  Let’s perform the rotation of ${60}^{\circ }$ about the point $A$.  When $P$ is the image of the point $C$, the triangle $ACP$ is equilateral and its angles are ${60}^{\circ }$.  Let $F$ be any inner point of the triangle $ABC$ and $Q$ its image in the rotation.  We infer that if the sides of the triangle $ABC$ are all seen from $F$ in the angle ${120}^{\circ }$, then the points $B$, $F$, $Q$, $P$ lie on the same line.

Generally, the triangles $APQ$ and $ACF$ are congruent, whence  $CF=QP$.  From the equilateral triangles we obtain:

$$AF+BF+CF=FQ+BF+QP=BFQP$$

Here, the right hand side is minimal when the points $B$, $F$, $Q$, $P$ are collinear, in which case

$\mathrm{\angle }CFA=\mathrm{\angle }PQA={\mathrm{\hspace{0.33em}180}}^{\circ }-\mathrm{\angle }AQF={\mathrm{\hspace{0.33em}120}}^{\circ },$

$\mathrm{\angle }AFB={\mathrm{\hspace{0.33em}180}}^{\circ }-\mathrm{\angle }QFA={\mathrm{\hspace{0.33em}120}}^{\circ },$

$\mathrm{\angle }BFC={\mathrm{\hspace{0.33em}360}}^{\circ }-{240}^{\circ }={\mathrm{\hspace{0.33em}120}}^{\circ }.$

Remark.  The point $F$ is called the Fermat point of the triangle $ABC$.