# Schooten theorem

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

 $AM=BM+CM$

holds.

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

## References

• 1 [Pritchard] Pritchard, Chris (ed.) The Changing Shape of Geometry : Celebrating a Century of Geometry and Geometry Teaching. Cambridge University Press, 2003.
Title Schooten theorem SchootenTheorem 2013-03-22 14:05:50 2013-03-22 14:05:50 mathcam (2727) mathcam (2727) 7 mathcam (2727) Theorem msc 51-00 Ptolemy’s theorem