# proof of Steiner’s theorem

Using $\alpha ,\beta ,\gamma ,\delta $ to denote angles as in the diagram at left, the sines law yields

$\frac{AB}{\mathrm{sin}(\gamma )}$ | $=$ | $\frac{AC}{\mathrm{sin}(\beta )}$ | (1) | ||

$\frac{NB}{\mathrm{sin}(\alpha +\delta )}$ | $=$ | $\frac{NA}{\mathrm{sin}(\beta )}$ | (2) | ||

$\frac{MC}{\mathrm{sin}(\alpha +\delta )}$ | $=$ | $\frac{MA}{\mathrm{sin}(\gamma )}$ | (3) | ||

$\frac{MB}{\mathrm{sin}(\alpha )}$ | $=$ | $\frac{MA}{\mathrm{sin}(\beta )}$ | (4) | ||

$\frac{NC}{\mathrm{sin}(\alpha )}$ | $=$ | $\frac{NA}{\mathrm{sin}(\gamma )}$ | (5) |

Dividing (2) and (3), and (4) by (5):

$$\frac{MA}{NA}\frac{NB}{MC}=\frac{\mathrm{sin}(\gamma )}{\mathrm{sin}(\beta )}=\frac{NA}{MA}\frac{MB}{NC}$$ |

and therefore

$$\frac{NB\cdot MB}{MC\cdot NC}=\frac{{\mathrm{sin}}^{2}(\gamma )}{{\mathrm{sin}}^{2}(\beta )}=\frac{A{B}^{2}}{A{C}^{2}}$$ |

by (1).

Title | proof of Steiner’s theorem |
---|---|

Canonical name | ProofOfSteinersTheorem |

Date of creation | 2013-03-22 13:48:06 |

Last modified on | 2013-03-22 13:48:06 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 5 |

Author | mathcam (2727) |

Entry type | Proof |

Classification | msc 51N20 |