Schooten theorem

Theorem: Let $A B C$ 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 $C B M$ 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
Canonical name	SchootenTheorem
Date of creation	2013-03-22 14:05:50
Last modified on	2013-03-22 14:05:50
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	7
Author	mathcam (2727)
Entry type	Theorem
Classification	msc 51-00
Synonym	Ptolemy’s theorem