# triangle mid-segment theorem

Theorem. The segment connecting the midpoints^{} of any two sides of a triangle is parallel^{} to the third side and is half as long.

Proof. In the triangle $ABC$, let ${A}^{\prime}$ be the midpoint of $AC$ and ${B}^{\prime}$ the midpoint of $BC$. Using the side-vectors $\overrightarrow{AC}$ and $\overrightarrow{CB}$ as a basis (http://planetmath.org/Basis) of the plane, we calculate the mid-segment ${A}^{\prime}{B}^{\prime}$ as a vector:

$$\overrightarrow{{A}^{\prime}{B}^{\prime}}=\overrightarrow{{A}^{\prime}C}+\overrightarrow{C{B}^{\prime}}=\frac{1}{2}\overrightarrow{AC}+\frac{1}{2}\overrightarrow{CB}=\frac{1}{2}(\overrightarrow{AC}+\overrightarrow{CB})=\frac{1}{2}\overrightarrow{AB}$$ |

The last expression indicates that the segment ${A}^{\prime}{B}^{\prime}$ is such as asserted.

Corollary (Varignon’s theorem). If one connects the midpoints of the of a quadrilateral^{}, one obtains a parallelogram^{}.

Title | triangle mid-segment theorem |

Canonical name | TriangleMidsegmentTheorem |

Date of creation | 2013-03-22 17:46:35 |

Last modified on | 2013-03-22 17:46:35 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 12 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 51M04 |

Classification | msc 51M25 |

Synonym | mid-segment theorem |

Related topic | MutualPositionsOfVectors |

Related topic | ParallelogramTheorems |

Related topic | MedianOfTrapezoid |

Related topic | CommonPointOfTriangleMedians |

Related topic | Grafix |

Related topic | SimonStevin |

Related topic | InterceptTheorem |