Triangle/Angle bisectors/Intersecting point/Fact/Proof

Due to fact, the angle bisector for ${\displaystyle {}A}$ consists of points that have the same distance to the line of the adjacent sides ${\displaystyle {}\mathbb {R} {\overrightarrow {AB}}}$ and ${\displaystyle {}\mathbb {R} {\overrightarrow {AC}}}$. Accordingly, the angle bisector for ${\displaystyle {}B}$ consists of points that have the same distance to the lines ${\displaystyle {}\mathbb {R} {\overrightarrow {BA}}}$ and ${\displaystyle {}\mathbb {R} {\overrightarrow {BC}}}$. Therefore, the intersecting point of these two angle bisectors has the same distance to all three sides.

In order to determine the coordinates, we write the angle bisector of ${\displaystyle {}A}$ as

${\displaystyle A+s{\left({\frac {\overrightarrow {AB}}{c}}+{\frac {\overrightarrow {AC}}{b}}\right)},}$

that is,

${\displaystyle {\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}}+s{\left({\frac {1}{c}}{\begin{pmatrix}b_{1}-a_{1}\\b_{2}-a_{2}\end{pmatrix}}+{\frac {1}{b}}{\begin{pmatrix}c_{1}-a_{1}\\c_{2}-a_{2}\end{pmatrix}}\right)}.}$

Equating with the angle bisector of ${\displaystyle {}B}$, we obtain

${\displaystyle {\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}}+s{\left({\frac {1}{c}}{\begin{pmatrix}b_{1}-a_{1}\\b_{2}-a_{2}\end{pmatrix}}+{\frac {1}{b}}{\begin{pmatrix}c_{1}-a_{1}\\c_{2}-a_{2}\end{pmatrix}}\right)}={\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}}+t{\left({\frac {1}{c}}{\begin{pmatrix}a_{1}-b_{1}\\a_{2}-b_{2}\end{pmatrix}}+{\frac {1}{a}}{\begin{pmatrix}c_{1}-b_{1}\\c_{2}-b_{2}\end{pmatrix}}\right)}\,.}$

The solution is given by

${\displaystyle {}s={\frac {bc}{a+b+c}}\,,}$

and

${\displaystyle {}t={\frac {ac}{a+b+c}}\,,}$

because this gives, after inserting, the symmetric expression

${\displaystyle {}{\begin{aligned}{\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}}+{\frac {bc}{a+b+c}}{\left({\frac {1}{c}}{\begin{pmatrix}b_{1}-a_{1}\\b_{2}-a_{2}\end{pmatrix}}+{\frac {1}{b}}{\begin{pmatrix}c_{1}-a_{1}\\c_{2}-a_{2}\end{pmatrix}}\right)}&={\frac {1}{a+b+c}}{\begin{pmatrix}a_{1}(a+b+c)+b(b_{1}-a_{1})+c(c_{1}-a_{1})\\a_{2}(a+b+c)+b(b_{2}-a_{2})+c(c_{2}-a_{2})\end{pmatrix}}\\&={\frac {1}{a+b+c}}{\begin{pmatrix}aa_{1}+bb_{1}+cc_{1}\\aa_{2}+bb_{2}+cc_{2}\end{pmatrix}}.\,\end{aligned}}}$

This is also the intersecting point of all three angle bisectors.