8.6: Incenter
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
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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.