# proof of Pythagorean theorem

This is a geometrical proof of Pythagorean theorem^{}. We begin with our
triangle:

$$\text{{xy}},(0,0);(20,0)**\mathrm{@}-;(20,10)**\mathrm{@}-;(0,0)**\mathrm{@}-,(10,-2)*a,(23,6)*b,(10,7)*c$$ |

Now we use the hypotenuse^{} as one side of a square:

$$\text{{xy}},(0,0);(20,0)**\mathrm{@}-;(20,10)**\mathrm{@}-;(0,0)**\mathrm{@}-;(-10,20)**\mathrm{@}-;(10,30)**\mathrm{@}-;(20,10)**\mathrm{@}-,(10,-2)*a,(23,6)*b,(10,7)*c$$ |

and draw in four more identical triangles

$$\text{{xy}},(0,0);(20,0)**\mathrm{@}-;(20,10)**\mathrm{@}-;(0,0)**\mathrm{@}-;(-10,20)**\mathrm{@}-;(10,30)**\mathrm{@}-;(20,10)**\mathrm{@}-;(20,30)**\mathrm{@}-;(-10,30)**\mathrm{@}-;(-10,0)**\mathrm{@}-;(0,0)**\mathrm{@}-,(10,-2)*a,(23,6)*b,(10,7)*c$$ |

Now for the proof. We have a large square, with each side of length $a+b$, which is subdivided into one smaller square and four triangles. The area of the large square must be equal to the combined area of the shapes it is made out of, so we have

${\left(a+b\right)}^{2}$ | $=$ | ${c}^{2}+4\left({\displaystyle \frac{1}{2}}ab\right)$ | |||

${a}^{2}+{b}^{2}+2ab$ | $=$ | ${c}^{2}+2ab$ | |||

${a}^{2}+{b}^{2}$ | $=$ | ${c}^{2}$ | (1) |

Title | proof of Pythagorean theorem |
---|---|

Canonical name | ProofOfPythagoreanTheorem |

Date of creation | 2013-03-22 11:56:36 |

Last modified on | 2013-03-22 11:56:36 |

Owner | drini (3) |

Last modified by | drini (3) |

Numerical id | 8 |

Author | drini (3) |

Entry type | Proof |

Classification | msc 51-00 |

Related topic | PythagorasTheorem |