Schooten theorem

Theorem: Let $ABC$ be a equilateral triangle. If $M$ is apoint on the circumscribed circle then the equality

$$AM=BM+CM$$

holds.

Proof: Let $B^{\prime }\in (MA)$ so that $M{B}^{\prime }=B^{\prime }B$. Because$\widehat{BMA}=\widehat{BCA}={60}^{\circ }$, the triangle $MB{B}^{\prime }$ isequilateral, so $B{B}^{\prime }=MB=M{B}^{\prime }$. Because $AB=BC,B{B}^{\prime }=BM$ and$\widehat{AB{B}^{\prime }}\equiv \widehat{MBC}$ we have that the triangles$AB{B}^{\prime }$ and $CBM$ are equivalent. Since $MC=A{B}^{\prime }$ we have that$AM=A{B}^{\prime }+B^{\prime }M=MC+MB$. $\mathrm{\square }$