在三角形中画一条角平分线

Question

我想从AB这个三角形的线段画一条角平分线，但我不知道如何才能让它不超出三角形。这是我的作品：

\documentclass{article}
\usepackage{tikz,amsmath,amssymb,tkz-euclide}
\begin{document}
    \begin{center}
        \begin{tikzpicture}
        \tkzDefPoint(0,0){A}
        \tkzLabelPoints[below,left](A)
        \tkzDefPoint(2.25,3.320718914){B}
        \tkzLabelPoints[above](B)
        \tkzDefPoint(6,0){C}
        \tkzLabelPoints[below,right](C)
        \tkzDrawSegment(A,B)
        \tkzDrawSegment(B,C)
        \tkzDrawSegment(A,C)
        \tkzDefLine[bisector](A,B,C)\tkzGetPoint{a}
        \tkzDrawSegment(B,a)
        \end{tikzpicture}
    \end{center}
\end{document}

多谢！

Answer 1

获取线段的交点Ba和AC

\tkzInterLL(A,C)(B,a) \tkzGetPoint{b}

然后画线段Bb

\documentclass{article}
\usepackage{tikz,amsmath,amssymb,tkz-euclide}
\begin{document}
    \begin{center}
        \begin{tikzpicture}
        \tkzDefPoint(0,0){A}
        \tkzLabelPoints[below,left](A)
        \tkzDefPoint(2.25,3.320718914){B}
        \tkzLabelPoints[above](B)
        \tkzDefPoint(6,0){C}
        \tkzLabelPoints[below,right](C)
        \tkzDrawSegments(A,B B,C A,C)
        \tkzDefLine[bisector](A,B,C)\tkzGetPoint{a}
        \tkzInterLL(A,C)(B,a) \tkzGetPoint{b}
        \tkzLabelPoints[below](b)
        \tkzDrawSegment(B,b)
        \end{tikzpicture}
    \end{center}
\end{document}