# some proofs for triangle theorems

In the following, only Euclidean geometry^{} is considered.

The sum of three angles $A$, $B$, and $C$ of a triangle is $A+B+C={180}^{\circ}$.

The following triangle shows how the angles can be found to make a half revolution, which equals ${180}^{\circ}$.

$\mathrm{\square}$

The area $A=rs$ where $s$ is the semiperimeter $s={\displaystyle \frac{a+b+c}{2}}$ and $r$ is the radius of the inscribed circle can be proven by creating the triangles $\mathrm{\u25b3}BAO$, $\mathrm{\u25b3}BCO$, and $\mathrm{\u25b3}ACO$ from the original triangle $\mathrm{\u25b3}ABC$, where $O$ is the center of the inscribed circle.

$\begin{array}{cc}\hfill {A}_{\mathrm{\u25b3}ABC}& ={A}_{\mathrm{\u25b3}ABO}+{A}_{\mathrm{\u25b3}BCO}+{A}_{\mathrm{\u25b3}ACO}\hfill \\ & \\ & ={\displaystyle \frac{rc}{2}}+{\displaystyle \frac{ra}{2}}+{\displaystyle \frac{rb}{2}}\hfill \\ & \\ & ={\displaystyle \frac{r(a+b+c)}{2}}\hfill \\ & \\ & =rs\hfill \end{array}$

$\mathrm{\square}$

Title | some proofs for triangle theorems |
---|---|

Canonical name | SomeProofsForTriangleTheorems |

Date of creation | 2013-03-22 14:03:55 |

Last modified on | 2013-03-22 14:03:55 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 13 |

Author | Wkbj79 (1863) |

Entry type | Proof |

Classification | msc 51-00 |