8.6: Incenter

Theorem $\PageIndex{1}$

The angle bisectors of any nondegenerate triangle intersect at one point.

The point of intersection of bisectors is called the incenter of the triangle; it is usually denoted by $I$. The point $I$ lies on the same distance from each side. In particular, it is the center of a circle tangent to each side of triangle. This circle is called the incircle and its radius is called the inradius of the triangle.

Proof

Let $\triangle ABC$ be a nondegenerate triangle.

Note that the points $B$ and $C$ lie on opposite sides of the bisector of $\angle BAC$. Hence this bisector intersects $[BC]$ at a point, say $A'$.

Analogously, there is $B' \in [AC]$ such that $(BB')$ bisects $\angle ABC$.

Applying Pasch's theorem (Theorem 3.4.1) twice for the triangles $AA'C$ and $BB'C$, we get that $[AA']$ and $[BB']$ intersect. Suppose that $I$ denotes the point of intersection.

Let $X, Y$, and $Z$ be the foot points of $I$ on $(BC)$, $(CA)$, and $(AB)$ respectively. Applying Proposition 8.5.1, we get that

$IY = IZ = IX.$

From the same lemma, we get that $I$ lies on the bisector or on the exterior bisector of $\angle BCA$.

The line $(CI)$ intersects $[BB']$, the points $B$ and $B'$ lie on opposite sides of $(CI)$. Therefore, the angles $ICB'$ and $ICB$ have opposite signs. Note that $\angle ICA = \angle ICB'$. Therefore, $(CI)$ cannot be the exterior bisector of $\angle BCA$. Hence the result.