Let **D** **ABC** be an acute triangle. Let ** l**
be the line which contains the angle bisector of

SinceIlies on the angle bisector of< BAC,Iis equidistant from the sides (ABandAC)of<BAC. (Angle Bisector Theorem) LetIFandIEbe perpendicular segments fromIto the sidesABandAC, respectively. Then,IF = IE.

Since

Ilies on the angle bisector of< ABC,Iis equidistant from the sides (ABandBC)of< ABC. (Angle Bisector Theorem)Let

IFandIDbe perpendicular segments fromIto the sidesABandBC, respectively. Then,IF = ID.

SinceIE = IFandIF = ID, thenIE = ID. (Transitive Property) Therefore,Iis equidistant to the sides (BCandAC)of< ACB.Imust also lie on the angle bisector of<ACB. (Converse of the Angle Bisector Theorem) Therefore, the three angle bisectors are concurrent at a pointIequidistant from the three sides. This point is called the. The incenter is the center of the inscribed circle ofincenterDABC.

Let **D** **ABC** be an obtuse triangle with **< ABC**
as an obtuse angle. All the conditions in the proof for an acute
triangle hold for an obtuse triangle.

Let **D ABC** be a right triangle
with **< ABC** as a right angle. All the conditions in the
proof for an acute triangle hold for an obtuse triangle. Additionally,
notice that **ID || AB** since both **AB** and **ID**
are perpendicular to **BC**. Also, **IF || BC** since both
**BC** and **IF** are perpendicular to **AB**.

The incenter is *always* located inside
a triangle since it lies within the interior of each of the three
angles.