# proof of Pythagorean theorem

Let $ABC$ be a right triangle^{} with hypotenuse^{} $BC$. Draw the height $AT$.

Using the right angles^{} $\mathrm{\angle}BAC$ and $\mathrm{\angle}ATB$ and the fact that the sum of angles on any triangle is ${180}^{\circ}$, it can be shown that

$\mathrm{\angle}BAT$ | $=$ | $\mathrm{\angle}ACT$ | ||

$\mathrm{\angle}TAC$ | $=$ | $\mathrm{\angle}CBA$ |

and therefore we have the following triangle similarities^{}:

$$\mathrm{\u25b3}ABC\sim \mathrm{\u25b3}TBA\sim \mathrm{\u25b3}TAC.$$ |

From those similarities, we have $\frac{AB}{BC}=\frac{TB}{BA}$ and thus $A{B}^{2}=BC\cdot TB$. Also $\frac{AC}{BC}=\frac{TC}{AC}$ and thus $A{C}^{2}=BC\cdot TC$. We have then

$$A{B}^{2}+A{C}^{2}=BC(BT+TC)=BC\cdot BC=B{C}^{2}$$ |

which concludes the proof.

Title | proof of Pythagorean theorem^{} |
---|---|

Canonical name | ProofOfPythagoreanTheorem1 |

Date of creation | 2013-03-22 12:48:39 |

Last modified on | 2013-03-22 12:48:39 |

Owner | drini (3) |

Last modified by | drini (3) |

Numerical id | 8 |

Author | drini (3) |

Entry type | Proof |

Classification | msc 51-00 |