# proof of pivot theorem

Let $\mathrm{\u25b3}ABC$ be a triangle, and let $D$, $E$, and $F$ be points
on $BC$, $CA$, and $AB$, respectively. The circumcircles^{} of
$\mathrm{\u25b3}AEF$ and $\mathrm{\u25b3}BFD$ intersect in $F$ and in another
point, which we call $P$. Then $AEPF$ and $BFPD$ are cyclic
quadrilaterals^{}, so

$$\mathrm{\angle}A+\mathrm{\angle}EPF=\pi $$ |

and

$$\mathrm{\angle}B+\mathrm{\angle}FPD=\pi $$ |

Combining this with $\mathrm{\angle}A+\mathrm{\angle}B+\mathrm{\angle}C=\pi $ and $\mathrm{\angle}EPF+\mathrm{\angle}FPD+\mathrm{\angle}DPE=2\pi $, we get

$$\mathrm{\angle}C+\mathrm{\angle}DPE=\pi .$$ |

This implies that $CDPE$ is a cyclic quadrilateral as well, so that $P$ lies on the circumcircle of $\mathrm{\u25b3}CDE$. Therefore, the circumcircles of the triangles $AEF$, $BFD$, and $CDE$ have a common point, $P$.

Title | proof of pivot theorem |
---|---|

Canonical name | ProofOfPivotTheorem |

Date of creation | 2013-03-22 13:12:37 |

Last modified on | 2013-03-22 13:12:37 |

Owner | pbruin (1001) |

Last modified by | pbruin (1001) |

Numerical id | 4 |

Author | pbruin (1001) |

Entry type | Proof |

Classification | msc 51M04 |