# common point of triangle medians

Theorem. The three medians (http://planetmath.org/Median) of a triangle intersect one another in one point, which divides each median in the ratio $2:1$.

Proof. Let the medians of a triangle $ABC$ be $AD$, $BE$ and $CF$. Any median vector is the arithmetic mean^{} of the side vectors emanating from the same vertex. Using vectors, let us form three ways all beginning from the vertex $A$, the first going simply $2/3$ of the median vector $\overrightarrow{AD}$ ( in the picture):

$\frac{2}{3}}\overrightarrow{AD}={\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{1}{2}}(\overrightarrow{AB}+\overrightarrow{AC})={\displaystyle \frac{1}{3}}(\overrightarrow{AB}+\overrightarrow{AC})$ | (1) |

The second way goes first the side vector $\overrightarrow{AB}$ and then $2/3$ of the median vector $\overrightarrow{BE}$ (green in the picture):

$\overrightarrow{AB}+{\displaystyle \frac{2}{3}}\overrightarrow{BE}=\overrightarrow{AB}+{\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{1}{2}}\left[-\overrightarrow{AB}+(\overrightarrow{AC}-\overrightarrow{AB})\right]={\displaystyle \frac{1}{3}}(\overrightarrow{AB}+\overrightarrow{AC})$ | (2) |

Similarly, the third way goes first the side vector $\overrightarrow{AC}$ and then $2/3$ of the median vector $\overrightarrow{CF}$ (red in the picture):

$\overrightarrow{AC}+{\displaystyle \frac{2}{3}}\overrightarrow{CF}=\overrightarrow{AC}+{\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{1}{2}}\left[-\overrightarrow{AC}+(\overrightarrow{AB}-\overrightarrow{AC})\right]={\displaystyle \frac{1}{3}}(\overrightarrow{AB}+\overrightarrow{AC})$ | (3) |

Thus the ways (2) and (3), where one goes from $A$ to another vertex and continues along the corresponding median $2/3$ of its length, lead to the point $M$ which is attained directly along $AD$. This means that all medians intersect in $M$. The distance^{} of $M$ from any vertex is $2/3$ of the corresponding median, and so the rest of the median is $1/3$ of its length, i.e. the ratio of the parts of any median is $2:1$.

Title | common point of triangle medians |
---|---|

Canonical name | CommonPointOfTriangleMedians |

Date of creation | 2013-03-22 17:46:54 |

Last modified on | 2013-03-22 17:46:54 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 8 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 51M04 |

Related topic | MutualPositionsOfVectors |

Related topic | ParallelogramPrinciple |

Related topic | DifferenceOfVectors |

Related topic | TriangleMidSegmentTheorem |

Related topic | LengthsOfTriangleMedians |