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Schooten theorem (Theorem)

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

$\displaystyle AM=BM+CM$
holds.

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

\includegraphics{schooten}

Bibliography

1
[Pritchard] Pritchard, Chris (ed.) The Changing Shape of Geometry : Celebrating a Century of Geometry and Geometry Teaching. Cambridge University Press, 2003.



"Schooten theorem" is owned by mathcam. [ full author list (2) | owner history (1) ]
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Other names:  Ptolemy's theorem
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Cross-references: equivalent, equilateral, triangle, proof, equality, circle, circumscribed, point, equilateral triangle

This is version 4 of Schooten theorem, born on 2003-12-08, modified 2006-01-12.
Object id is 5482, canonical name is SchootenTheorem.
Accessed 2688 times total.

Classification:
AMS MSC: 51-00 (Geometry :: General reference works )
Pending Errata and Addenda
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