proof of triangle incenter

In ABC and construct bisectorsMathworldPlanetmath of the angles at A and C, intersecting at O11Note that the angle bisectorsMathworldPlanetmath must intersect by Euclid’s Postulate 5, which states that “if a straight line falling on two straight lines makes the interior anglesMathworldPlanetmath on the same side less than two right anglesMathworldPlanetmathPlanetmath, 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 incenterMathworldPlanetmath of ABC.


Drop perpendiculars from O to each of the three sides, intersecting the sides in D, E, and F. Clearly, by AAS, CODCOE and also AOEAOF. Thus FO=EO=DO. It follows that O is the incenter of ABC since its distanceMathworldPlanetmath from all three sides is equal.

Also, since FO=DO we see that BOF and BOD are right triangles with two equal sides, so by SSA (which is applicable for right triangles), BOFBOD. Thus BO bisects 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