intersection point of bisector in triangle

Question

\usetikzlibrary{calc,angles,positioning,intersections,quotes,decorations.markings}
\usepackage{tkz-euclide}
\usetkzobj{all}
\usepackage{tikz}

\usepackage[utf8]{inputenc}
\usepackage[english,greek]{babel} %βαζω την ελληνική γλώσσα...όλα θα γινουν ελληνικα αν δεν χρησιμοποιήσω το \en
\usepackage{alphabeta}
\usepackage{multicol}
\newcommand{\en}{\selectlanguage{english}}
\newcommand{\gr}{\selectlanguage{greek}}
\author{\textlatin{}}
\usepackage{array,amsmath}
\usepackage{amsthm}
\usepackage{lipsum}
\usepackage[T1]{fontenc}
\usepackage{smartdiagram}
\begin{tikzpicture}\hspace{-2mm}
    \coordinate (A) at (5,0);
    \coordinate (B) at (4,-3);
    \coordinate (G) at (8,-3);

    \filldraw[black] (A) circle (0.1pt) node[anchor=south] {A};
    \filldraw[black] (B) circle (0.1pt) node[anchor=north] {B};
    \filldraw[black] (G) circle (0.1pt) node[anchor=north] {G};

    \draw [black,thick] (A)--(B)--(G)-- cycle;

    \tkzDrawBisector[color=green](A,B,G)
    \end{tikzpicture} 

This is a part of my code. I draw a triangle ABG and find the bisector. I want to know how I can find the intersection point of the bisector and the AG line without coordinating a point myself. I mean, is there a way to draw a line from B to intersection point of bisector with AG??

Answer 1

Use \tkzGetPoint for the point of intersection.

\tkzDrawBisector[color=green](A,B,G)\tkzGetPoint{C}
        \filldraw (C) circle (2pt);

Answer 2

\begin{tikzpicture}\hspace{-2mm}
    \coordinate (A) at (5,0);
    \coordinate (B) at (4,-3);
    \coordinate (G) at (8,-3);
    \draw [black,thick] (A)node[anchor=south]{A}--(B)node[anchor=north]{B}--(G)node[anchor=north]{G}-- cycle;

    \tkzDefLine[bisector](A,B,G) \tkzGetPoint{x}
    \tkzInterLL(B,x)(A,G) \tkzGetPoint{I}  %intersection of lines Bx and AG
    \draw (B)--(I);
    \tkzDrawPoints[color=black](A,B,G,I)
    \tkzLabelPoints[above right](I)


    \end{tikzpicture} 

Answer 3

You are already loading the calc library, why not using it?

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
    \coordinate (A) at (5,0);
    \coordinate (B) at (4,-3);
    \coordinate (G) at (8,-3);

    \filldraw[black] (A) circle (0.1pt) node[anchor=south] {A};
    \filldraw[black] (B) circle (0.1pt) node[anchor=north] {B};
    \filldraw[black] (G) circle (0.1pt) node[anchor=north] {G};

    \draw [black,thick] (A)--(B)--(G)-- cycle;
    \draw[blue]   (B) -- ($(A)!(B)!(G)$) node[circle,fill,inner sep=1pt,blue]{};
\end{tikzpicture} 
\end{document}