Apparently, the incentre of a triangle, if it lies on a median, divides it in the ratio $$\frac{BD}{DF} = \frac{AB+BC}{AC} $$ as per this figure(where $BF$ is the median) : 
Proving it for isosceles triangles is easy enough (as the incentre lies on the median for all isosceles triangles), but I can't prove it generally. I obtained an equation when I substituted the coordinates of the incentre in the equation for the median, but can't find a way to use it in the distance formulas to prove the above theorem.
Could anyone help point out how to prove it for the general case?
EDIT: the equation I got, using coordinates $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ for the vertices, is $x_2y_1-x_3y_1-x_1y_2+x_1y_3=x_2y_3-x_3y_2$. (The values of the sides of the triangle, which come with the coordinates of the incentre, vanish by the end.)
Best Answer
$\text{Note:}$ Here I have used angle bisector theorem for triangles
If incenter lies on median , then that median is one of the angle bisectors of triangle
$\displaystyle\frac{AB}{AF}=\frac{BC}{CF}\implies AB=BC$
as $AF=FC$ as $BF$ is median proving this is possible in case of isosceles triangle only
Make another angle bisector $ADE$ and $CDP$ and then you have
$\displaystyle\frac{AF}{AB}=\frac{FD}{DB}=\frac{CF}{CB}$
Therefore $2\displaystyle\frac{BD}{DF}=\frac{CB+AB}{AF}\implies \frac{BD}{DF}=\frac{CB+AB}{2AF}=\frac{CB+AB}{AC} $