# proof of triangle incenter

In $\mathrm{\u25b3}ABC$ and construct bisectors^{} of the angles at $A$ and $C$, intersecting at $O$^{1}^{1}Note that the angle bisectors^{} must intersect by Euclid’s Postulate 5, which states that “if a straight line falling on two straight lines makes the interior angles^{} on the same side less than two right angles^{}, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.” They must meet inside the triangle by considering which side of $AB$ and $CB$ they fall on. Draw $BO$. We show that $BO$ bisects the angle at $B$, and that $O$ is in fact the incenter^{} of $\mathrm{\u25b3}ABC$.

Drop perpendiculars from $O$ to each of the three sides, intersecting the sides in $D$, $E$, and $F$. Clearly, by AAS, $\mathrm{\u25b3}COD\cong \mathrm{\u25b3}COE$ and also $\mathrm{\u25b3}AOE\cong \mathrm{\u25b3}AOF$. Thus $FO=EO=DO$. It follows that $O$ is the incenter of $\mathrm{\u25b3}ABC$ since its distance^{} from all three sides is equal.

Also, since $FO=DO$ we see that $\mathrm{\u25b3}BOF$ and $\mathrm{\u25b3}BOD$ are right triangles with two equal sides, so by SSA (which is applicable for right triangles), $\mathrm{\u25b3}BOF\cong \mathrm{\u25b3}BOD$. Thus $BO$ bisects $\mathrm{\angle}ABC$.

Title | proof of triangle incenter |
---|---|

Canonical name | ProofOfTriangleIncenter |

Date of creation | 2013-03-22 17:12:26 |

Last modified on | 2013-03-22 17:12:26 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 6 |

Author | rm50 (10146) |

Entry type | Proof |

Classification | msc 51M99 |