Schooten theorem
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 $M{B}^{\prime}={B}^{\prime}B$. Because $\widehat{BMA}=\widehat{BCA}={60}^{\circ}$, the triangle^{} $MB{B}^{\prime}$ is equilateral, 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 $AM=A{B}^{\prime}+{B}^{\prime}M=MC+MB$. $\mathrm{\square}$
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 |