Let $BD$ be the internal angle bisector of $\Delta ABC$ with $D$ on $AC$. The incentre of $\Delta ABC$ is $(0,4)$ and $D$ is at $(1,3)$ 
Let $BD$ be the internal angle bisector of $\Delta ABC$ with $D$ on $AC$. The incentre of $\Delta ABC$ is $(0,4)$ and $D$ is at $(1,3)$. If $a,b,c$ are in arithmetic progression, find the point $B$. ($a,b,c$ represents the sides $BC,AC,AB$).

Incentre of a triangle lies on the angle bisector. I can get the parametric point of $B$.
The line joining incentre and $D$ is $$y-4=\frac{4-3}{0-1}(x-0)$$
$$y-4=-x$$
$$y=-x+4$$
Hence $B=(h,4-h)$
I could not solve the problem further. How should I use the information about $a,b,c$?

From the Bisector Theorem we have
\begin{align*}
\frac{|AD|}{|DC|}&=\frac{|AB|}{|BC|}=\frac{c}{a}\qquad\implies\qquad |AD|=\frac{c}a|DC|=\frac{c}a\left(b-|AD|\right)
\end{align*}
By solving for $|AD|$ the last equation we get
$$|AD|=\frac{bc}{a+c}$$
Since $a,b$ and $c$ are in arithmetic progression it follows that $a+c=2b$, then
$$|AD|=\frac{bc}{2b}=\frac{c}{2}\tag{1}$$
Let $I$ be the incenter of the triangle $ABC$, from the Bisector Theorem again we get
$$\frac{|BI|}{|ID|}=\frac{|AB|}{|AD|}=\frac{c}{c/2}=2\qquad\implies\qquad |BI|=2|ID|=2\sqrt2$$
Which means
\begin{align*}
|BI|&=2\sqrt{2}\\
\sqrt{h^2+(4-(4-h))^2}&=2\sqrt{2}\\
\sqrt{h^2+h^2}&=2\sqrt{2}\\
|h|\sqrt{2}&=2\sqrt{2}\\
\end{align*}
It follows that $h=-2$ in order to make sure that $I$ is inside the triangle $ABC$, then the coordinates of $B$ are $(-2,6)$.

