# Euler line proof

Let $O$ the circumcenter^{} of $\mathrm{\u25b3}ABC$ and $G$ its centroid. Extend $OG$ until a point $P$ such that $OG/GP=1/2$. We’ll prove that $P$ is the orthocenter^{} $H$.

Draw the median $A{A}^{\prime}$ where ${A}^{\prime}$ is the midpoint^{} of $BC$. Triangles $OG{A}^{\prime}$ and $PGA$ are similar^{}, since $GP=2GO$, $AG=2{A}^{\prime}G$ and $\mathrm{\angle}OG{A}^{\prime}=\mathrm{\angle}PGA$. Then $\mathrm{\angle}O{A}^{\prime}G=\mathrm{\angle}PGA$ and $O{A}^{\prime}\parallel AP$. But $O{A}^{\prime}\u27c2BC$ so $AP\u27c2BC$, that is, $AP$ is a height of the triangle.

Repeating the same argument for the other medians proves that $P$ lies on the three heights and therefore it must be the orthocenter $H$.

The ratio is $OG/GH=1/2$ since we constructed it that way.

Title | Euler line proof |

Canonical name | EulerLineProof |

Date of creation | 2013-03-22 11:44:29 |

Last modified on | 2013-03-22 11:44:29 |

Owner | drini (3) |

Last modified by | drini (3) |

Numerical id | 15 |

Author | drini (3) |

Entry type | Proof |

Classification | msc 51M99 |

Classification | msc 55U10 |

Classification | msc 18E30 |

Classification | msc 18-00 |

Classification | msc 55U35 |

Classification | msc 46-01 |

Classification | msc 47B25 |

Classification | msc 81-01 |

Related topic | EulerLine |