# intersection point of bisector in triangle

\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??

Use \tkzGetPoint for the point of intersection.

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


\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}


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}