Lighthouse at Fermat Points

The problem reminds of the Lighthouse theorem due to R. K. Guy, but the relation is tenuous. The proof below invokes homothety but does not see to employ Menelaus' theorem.

The solution draws on the following diagram:

Let $AB'C$ be one of the Napoleon's triangles so that Fermat point $F$ lies on $BB'.$ Let $\Delta IHA_1$ be a homothetic image of $\Delta AB'C,$ with the homothety centered at $B.$ Then $I\in AB,$ $H\in BB',$ $IA_{1}\parallel AC,$ $HA_{1}\parallel B'C,$ $IH\parallel AB',$ and $\Delta IHA_1$ is equilateral.

Let $P$ denote the intersection of $A_1B_1$ with $AB.$ We are going to show that $P=C_1.$

For a start, since $IA_{1}\parallel AB_1,$ triangles $A_{1}IP$ and $B_{1}AP$ are similar, implying

$\displaystyle\frac{A_{1}I}{B_{1}A}=\frac{A_{1}P}{B_{1}P}.$

From $\Delta IHA_1$ being equilateral and $IA_{1}\parallel AB_{1}$ we deduce that

$\angle (\overrightarrow{A_{1}H},\overrightarrow{B_{1}A})=\angle (\overrightarrow{HF},\overrightarrow{AF})=120^\circ,$

and, from here, $\angle FHA_{1}=\angle FAB_1.$ By the construction, $\angle A_{1}FB_{1}=120^\circ$ and, from the property of the Fermat points, $\angle AFB=120^\circ,$ implying $\angle B_{1}FA=\angle A_{1}FH,$ which makes triangles $A_{1}HF$ and $B_{1}AF$ similar:

$\displaystyle\frac{A_{1}H}{B_{1}A}=\frac{FA_{1}}{FB_{1}}.$

Combining the two proportions and bearing in mind that $A_{1}I=A_{1}H,$ we get

$\displaystyle\frac{A_{1}P}{B_{1}P}=\frac{FA_{1}}{FB_{1}}.$

Referring to a property of angle bisectors, this implies that $HP$ is the bisector of $\angle A_{1}FB_{1},$

which exactly means that $P=C_{1},$ as required.

This is Theorem 1 from a paper by Dao Thanh Oai announced at the CutTheKnotMath facebook page.