Congruence of angles with vertex at Fermat-Torricelli point through any two vertices

Question

At the following website, $\angle{ARF} \cong \angle{ABF}$ is asserted.

https://en.wikipedia.org/wiki/Fermat_point#Construction

(See section labeled Location of X(13).) The Inscribed-Angle Theorem is tacitly used. $A$, $B$, and $R$ are, by definition, points on the same circle. Why is F another point on the circle?

Answer 1

Equipped with which facts do you want to start? The angles are supplementary: $\angle ARB = 60^{\circ}$ and $\angle AFB = 120^{\circ}$.

For a quadrilateral, when an opposing pair of angles add up to $\pi$, it is cyclic.

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Assume you agree to start with the definition of a Fermat point as constructing equilateral triangles for each of the sides.

In the wikipedia figure shown below, compare the red $\color{red}{\triangle ARC}$ and the blue $\color{blue}{\triangle ABQ}$.

We have the short side matching: $\color{red}{\overline{AR}} = \color{blue}{\overline{AB}}$ because of the equilateral triangle of the side $\overline{AB}$. Similarly, we have the long side matching $\color{red}{\overline{AC}} = \color{blue}{\overline{AQ}}$ due to the equilateral triangle of the side $\overline{AC}$.

Further, we have $\color{red}{\angle RAC} = \frac{\pi}3 + \angle BAC = \color{blue}{\angle BAQ}$

Thus by side-angle-side we have the congruence between $\color{red}{\triangle ARC} \cong \color{blue}{\triangle ABQ}$

Now, since the two triangles share a vertex, point $A$, we see that $\color{blue}{\triangle ABQ}$ can be obtained by rotating $\color{red}{\triangle ARC}$, counter-clockwise by $\frac{\pi}3$ (or equivalently clockwise by $\frac{2\pi}3$.

That is, $\angle RFB = \frac{\pi}3$, which is half of our desired $\angle AFB$.

The other half comes from the same analysis on $\triangle ABP \cong \triangle RBC$, establishing the $\frac{\pi}3$ rotational relation between $\overline{AP}$ and $\overline{RC}$.