Let points A,B,C be the vertices of the triangle and vertex B be also a vertex of angle we want to divide.
Inspired by construction I derived following way to write equation for angle bisector :-
- Write equation for the lines including rays of triangle
- On the one of the rays choose point D
- Write equation of circle with center at B and radius BD
- Let point E be intersection of circle and other ray of the angle (ray on which don't lies point D)
- Write equation for line through points D and E
- Write equation for line perpendicular to DE and passing through the point B
Steps 5. and 6. can be combined into one if we know equation for perpendicular line
Problem is in step 4. We write system of equations from which we get two points E
but only one of them will give us equation of interior angle
and how to choose correct one
Let's perform some calculations. Warning: they are quite lengthy.
$$l_{AB}:y-y_{B}=\frac{y_{B}-y_{A}}{x_{B}-x_{A}}\left(x-x_{B}\right)\\ l_{AB}:\left(y-y_{B}\right)\left(x_{B}-x_{A}\right)=\left(y_{B}-y_{A}\right)\left(x-x_{B}\right)\\ l_{AB}:\left(y_{B}-y_{A}\right)\left(x-x_{B}\right)-\left(y-y_{B}\right)\left(x_{B}-x_{A}\right)=0\\ l_{AB}:\left(y_{B}-y_{A}\right)x-\left(x_{B}-x_{A}\right)y-x_{B}\left(y_{B}-y_{A}\right)+y_{B}\left(x_{B}-x_{A}\right)=0 $$ $$ l_{BC}:y-y_{C}=\frac{y_{C}-y_{B}}{x_{C}-x_{B}}\left(x-x_{C}\right)\\ l_{BC}:\left(y-y_{C}\right)\left(x_{C}-x_{B}\right)=\left(y_{C}-y_{B}\right)\left(x-x_{C}\right)\\ l_{BC}:\left(y_{C}-y_{B}\right)\left(x-x_{C}\right)-\left(x_{C}-x_{B}\right)\left(y-y_{C}\right)=0\\ l_{BC}:\left(y_{C}-y_{B}\right)x-\left(x_{C}-x_{B}\right)y-x_{C}\left(y_{C}-y_{B}\right)+y_{B}\left(x_{C}-x_{B}\right)=0 $$ $$ l_{AC}:y-y_{C}=\frac{y_{C}-y_{A}}{x_{C}-x_{A}}\left(x-x_{C}\right)\\ l_{AC}:\left(x_{C}-x_{A}\right)\left(y-y_{C}\right)=\left(y_{C}-y_{A}\right)\left(x-x_{C}\right)\\ l_{AC}:\left(y_{C}-y_{A}\right)\left(x-x_{C}\right)-\left(x_{C}-x_{A}\right)\left(y-y_{C}\right)=0\\ l_{AC}:\left(y_{C}-y_{A}\right)x-\left(x_{C}-x_{A}\right)y-x_{C}\left(y_{C}-y_{A}\right)+y_{C}\left(x_{C}-x_{A}\right)=0 $$
Let general equation of $l_{AB}$ be
$$A_{1}x+B_{1}y+C_{1}=0$$ Let general equation of $l_{BC}$ be
$$A_{2}x+B_{2}y+C_{2}=0$$ Let general equation of $l_{AC}$ be
$$A_{3}x+B_{3}y+C_{3}=0$$
Let's write bisector equations for vertices A and B.
Begin with vertex B. For simplicity, assume that our point D is equal to point A.
$$|BA|=\sqrt{\left(x_{A}-x_{B}\right)^2+\left(y_{A}-y_{B}\right)^2}\\ |BA|=\sqrt{A_{1}^2+B_{1}^2}\\ |BA|=c\\ |CB|=\sqrt{\left(x_{B}-x_{C}\right)^2+\left(y_{B}-y_{C}\right)^2}\\ |CB|=\sqrt{A_{2}^2+B_{2}^2}\\ |CB|=a $$
From step 4. we have system of equations
$$ \left \{ \begin{align} \left(x-x_{B}\right)^2+\left(y-y_{B}\right)^2=c^2 \\ A_{2}x+B_{2}y+C_{2}=0 \end{align} \right. $$ $$ \left \{ \begin{align} \left(x-x_{B}\right)^2+\left(y-y_{B}\right)^2=c^2 \\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} \left(x-x_{B}\right)^2+\left(-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right)-y_{B}\right)^2=c^2 \\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} \left(x-x_{B}\right)^2+\left(\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right)+y_{B}\right)^2=c^2 \\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} \left(x-x_{B}\right)^2+\left(\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right)-\frac{1}{B_{2}}\left(A_{2}x_{B}+C_{2}\right)\right)^2=c^2 \\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} \left(x-x_{B}\right)^2+\left(\frac{A_{2}}{B_{2}}x-\frac{A_{2}}{B_{2}}x_{B}\right)^2=c^2 \\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} \left(x-x_{B}\right)^2+\left(\frac{A_{2}}{B_{2}}\left(x-x_{B}\right)\right)^2=c^2 \\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} \left(x-x_{B}\right)^2+\frac{A_{2}^2}{B_{2}^2}\left(x-x_{B}\right)^2=c^2 \\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} \left(1+\frac{A_{2}^2}{B_{2}^2}\right)\left(x-x_{B}\right)^2=c^2 \\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} \frac{B_{2}^2+A_{2}^2}{B_{2}^2}\left(x-x_{B}\right)^2=c^2 \\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} \frac{a^2}{B_{2}^2}\left(x-x_{B}\right)^2=c^2 \\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} \left(x-x_{B}\right)^2=B_{2}^2\cdot\frac{c^2}{a^2} \\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} \left(x-x_{B}\right)^2-B_{2}^2\cdot\frac{c^2}{a^2}=0\\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} \left(x-x_{B}-B_{2}\cdot\frac{c}{a}\right)\left(x-x_{B}+B_{2}\cdot\frac{c}{a}\right)=0\\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} x=x_{B}\mp B_{2}\cdot\frac{c}{a} \\ y=-\frac{1}{B_{2}}\left(A_{2}x+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} x=x_{B}\mp B_{2}\cdot\frac{c}{a} \\ y=-\frac{1}{B_{2}}\left(A_{2}\left(x_{B}\mp B_{2}\cdot\frac{c}{a}\right)+C_{2}\right) \end{align} \right. \\ $$ $$ \left \{ \begin{align} x=x_{B}\mp B_{2}\cdot\frac{c}{a} \\ y=-\frac{1}{B_{2}}\left(A_{2}x_{B}+C_{2}\right)\pm \frac{1}{B_{2}}A_{2}B_{2}\cdot\frac{c}{a} \end{align} \right. \\ $$ $$ \left \{ \begin{align} x=x_{B}\mp B_{2}\cdot\frac{c}{a} \\ y=y_{B}\pm A_{2}\cdot\frac{c}{a} \end{align} \right. \\ $$
Now check which one is the point we want to choose
$$\left(A_{1}x_{E}+B_{1}y_{E}+C_{1}\right)\left(A_{1}x_{C}+B_{1}y_{C}+C_{1}\right)>0\\ \left(A_{1}\left(x_{B}\mp B_{2}\cdot\frac{c}{a}\right)+B_{1}\left(y_{B}\pm A_{2}\cdot\frac{c}{a}\right)+C_{1}\right)\left(A_{1}\left(\left(x_{C}-x_{B}\right)+x_{B}\right)+B_{1}\left(\left(y_{C}-y_{B}\right)+y_{B}\right)+C_{1}\right)>0\\ \left(\mp A_{1}B_{2}\cdot\frac{c}{a}\pm B_{1}A_{2}\cdot\frac{c}{a}+A_{1}x_{B}+B_{1}y_{B}+C_{1}\right)\left(A_{1}\left(-B_{2}+x_{B}\right)+B_{1}\left(A_{2}+y_{B}\right)+C_{1}\right)>0\\ \left(\mp A_{1}B_{2}\cdot\frac{c}{a}\pm B_{1}A_{2}\cdot\frac{c}{a}\right)\left(-A_{1}B_{2}+B_{1}A_{2}+A_{1}x_{B}+B_{1}x_{B}+C_{1}\right) > 0\\ \frac{c}{a}\cdot\left(\mp A_{1}B_{2}\pm B_{1}A_{2}\right)\left(-A_{1}B_{2}+B_{1}A_{2}\right) > 0 $$
Now it should be clear which point should we choose
$$ \left \{ \begin{align} x_{E}=x_{B}-B_{2}\cdot\frac{c}{a} \\ y_{E}=y_{B}+A_{2}\cdot\frac{c}{a} \end{align} \right. \\ $$
Let's calculate equation of line perpendicular to DE passing through point B
$$ y-y_{B}=-\frac{\left(x_{B}-B_{2}\cdot\frac{c}{a}\right)-x_{A}}{\left(y_{B}+A_{2}\cdot\frac{c}{a}\right)-y_{A}}\left(x-x_{B}\right)\\ y-y_{B}=-\frac{\left(\left(x_{B}-x_{A}\right)-B_{2}\cdot\frac{c}{a}\right)}{\left(\left(y_{B}-y_{A}\right)+A_{2}\cdot\frac{c}{a}\right)}\left(x-x_{B}\right)\\ y-y_{B}=-\frac{-B_{1}-B_{2}\cdot\frac{c}{a}}{A_{1}+A_{2}\cdot\frac{c}{a}}\left(x-x_{B}\right)\\ y-y_{B}=\frac{B_{1}a+B_{2}c}{A_{1}a+A_{2}c}\left(x-x_{B}\right)\\ \left(A_{1}a+A_{2}c\right)\left(y-y_{B}\right)=\left(B_{1}a+B_{2}c\right)\left(x-x_{B}\right)\\ \left(B_{1}a+B_{2}c\right)\left(x-x_{B}\right)-\left(A_{1}a+A_{2}c\right)\left(y-y_{B}\right)=0\\ \left(B_{1}a+B_{2}c\right)x-\left(A_{1}a+A_{2}c\right)y-x_{B}\left(B_{1}a+B_{2}c\right)+y_{B}\left(A_{1}a+A_{2}c\right)=0 $$
It looks good, but when i performed the same calculation for vertex A and tried to solve system of equations, i could not simplify the solution to this system of equations (to show that solution indeed gives the coordinates of the center of inscribed circle).
I will not present all calculations for vertex A because they are almost the same as for vertex B
$$ |AC|=\sqrt{\left(x_{C}-x_{A}\right)^2+\left(y_{C}-y_{A}\right)^2}\\ |AC|=\sqrt{A_{3}^2+B_{3}^2}\\ |AC|=b $$
System of equations from step 4. look as follows
$$ \left \{ \begin{align} \left(x-x_{A}\right)^2+\left(y-y_{A}\right)^2=c^2 \\ A_{3}x+B_{3}y+C_{3}=0 \end{align} \right. $$
Solution to this equation is
$$ \left \{ \begin{align} x=x_{A} \mp B_{3}\cdot\frac{c}{b}\\ y=y_{A} \pm A_{3}\cdot\frac{c}{b} \end{align} \right. $$
Now check which one is the point we want to choose
$$ \left(A_{1}x_{E}+B_{1}y_{E}+C_{1}\right)\left(A_{1}x_{C}+B_{1}y_{C}+C_{1}\right)>0\\ \frac{c}{b}\cdot\left(\mp A_{1}B_{3}\pm B_{1}A_{3}\right)\left(-A_{1}B_{3}+B_{1}A_{3}\right)>0 $$
Now it should be clear which point should we choose
$$ \left \{ \begin{align} x_{E}=x_{A} - B_{3}\cdot\frac{c}{b}\\ y_{E}=y_{A} + A_{3}\cdot\frac{c}{b} \end{align} \right. $$
Now let's calculate the equation of perpendicular to DE and passing through point A where D is chosen to be equal to B
$$ y-y_{A}=-\frac{\left(x_{A}-B_{3}\cdot\frac{c}{b}\right)-x_{B}}{\left(y_{A}+A_{3}\frac{c}{b}\right)-y_{B}}\left(x-x_{A}\right)\\ y-y_{A}=-\frac{\left(\left(x_{A}-x_{B}\right)-B_{3}\cdot\frac{c}{b}\right)}{\left(\left(y_{A}-y_{B}\right)+A_{3}\cdot\frac{c}{b}\right)}\left(x-x_{A}\right)\\ y-y_{A}=-\frac{B_{1}-B_{3}\cdot\frac{c}{b}}{-A_{1}+A_{3}\cdot\frac{c}{b}}\left(x-x_{A}\right)\\ y-y_{A}=\frac{B_{1}-B_{3}\cdot\frac{c}{b}}{A_{1}-A_{3}\cdot\frac{c}{b}}\left(x-x_{A}\right)\\ y-y_{A}=\frac{B_{1}b-B_{3}c}{A_{1}b-A_{3}c}\left(x-x_{A}\right)\\ \left(A_{1}b-A_{3}c\right)\left(y-y_{A}\right)=\left(B_{1}b-B_{3}c\right)\left(x-x_{A}\right)\\ \left(B_{1}b-B_{3}c\right)\left(x-x_{A}\right)-\left(A_{1}b-A_{3}c\right)\left(y-y_{A}\right)=0\\ \left(B_{1}b-B_{3}c\right)x-\left(A_{1}b-A_{3}c\right)y-x_{A}\left(B_{1}b-B_{3}c\right)+y_{A}\left(A_{1}b-A_{3}c\right)=0 $$
Now we have system of equations
$$ \left \{ \begin{align} \left(B_{1}b-B_{3}c\right)x-\left(A_{1}b-A_{3}c\right)y=x_{A}\left(B_{1}b-B_{3}c\right)-y_{A}\left(A_{1}b-A_{3}c\right)\\ \left(B_{1}a+B_{2}c\right)x-\left(A_{1}a+A_{2}c\right)y=x_{B}\left(B_{1}a+B_{2}c\right)-y_{B}\left(A_{1}a+A_{2}c\right) \end{align} \right. $$
Now solution to this equation in matrix form is
$$ \begin{bmatrix}\left(B_{1}b-B_{3}c\right) & -\left(A_{1}b-A_{3}c\right) \\\left(B_{1}a+B_{2}c\right) & -\left(A_{1}a+A_{2}c\right) \end{bmatrix}^{-1}\cdot\begin{bmatrix}x_{A}\left(B_{1}b-B_{3}c\right)-y_{A}\left(A_{1}b-A_{3}c\right) \\x_{B}\left(B_{1}a+B_{2}c\right)-y_{B}\left(A_{1}a+A_{2}c\right) \end{bmatrix} $$
I don't know how can I simplify this solution
(It should be the coordinates of center of inscribed circle)