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# 8.6: Incenter

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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.