The figure below represents the triangle and lines bisectors of
its sides.

The equation of $AB$ is$3x-2y+c=0$.

As it passes through$A(3,2),$ the value of $c=-5$.

Thus, $3x-2y-5=0$.

The coordinates of $D$ is $(1,-1)$ as $D$ is the mid-point of $AB$.

Consider the coordinates of $B$ are $(\alpha ,\beta ).$

$\frac{3+\alpha }{2}=1, \frac{2+\beta }{2}=-1$

$\alpha =-1, \beta =-4$

$B(-1,-4)$

Similarly, the equation of $AC$ is $2x-y+c=0$.

As it passes through $(3,2),$ the value of $c=-4$.

Thus,$2x-y-4=0$.

$E$ is a point of intersection of $x+2y-2=0$ and$2x-y-4=0$.

$E=(2,0)$ and $E$ is the midpoint of $AC$.

The coordinates of $C\left({\alpha }^{\text{'}},{\beta }^{\text{'}}\right)$ are,

$\frac{3+{\alpha }^{\text{'}}}{2}=2, \frac{2+{\beta }^{\text{'}}}{2}=0$

${\alpha }^{\text{'}}=1, {\beta }^{\text{'}}=-2$

$C(1,-2)$

The equation of $BC$ is,

$y+2=\left(\frac{-2+4}{1+1}\right)\left(x-1\right)$

Further simplification,