# alternate proof of parallelogram law

Proof of this is simple, given the cosine law:

$${c}^{2}={a}^{2}+{b}^{2}-2ab\mathrm{cos}\varphi $$ |

where $a$, $b$, and $c$ are the lengths of the sides of the triangle, and angle $\varphi $ is the corner angle opposite the side of length $c$.

Let us define the largest interior angles^{} as angle $\theta $.
Applying this to the parallelogram^{}, we find that

${d}_{1}^{2}$ | $=$ | ${u}^{2}+{v}^{2}-2uv\mathrm{cos}\theta $ | ||

${d}_{2}^{2}$ | $=$ | ${u}^{2}+{v}^{2}-2uv\mathrm{cos}\left(\pi -\theta \right)$ |

Knowing that

$$\mathrm{cos}\left(\pi -\theta \right)=-\mathrm{cos}\theta $$ |

we can add the two expressions together, and find ourselves with

${d}_{1}^{2}+{d}_{2}^{2}$ | $=$ | $2{u}^{2}+2{v}^{2}-2uv\mathrm{cos}\theta +2uv\mathrm{cos}\theta $ | ||

${d}_{1}^{2}+{d}_{2}^{2}$ | $=$ | $2{u}^{2}+2{v}^{2}$ |

which is the theorem we set out to prove.

Title | alternate proof of parallelogram law |
---|---|

Canonical name | AlternateProofOfParallelogramLaw |

Date of creation | 2013-03-22 12:43:52 |

Last modified on | 2013-03-22 12:43:52 |

Owner | drini (3) |

Last modified by | drini (3) |

Numerical id | 5 |

Author | drini (3) |

Entry type | Proof |

Classification | msc 51-00 |

Related topic | ProofOfParallelogramLaw2 |