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Mathematics LibreTexts

8.6: Incenter

  • Page ID
    23631
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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

    截屏2021-02-18 上午10.36.14.png

    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.