$$\left(\frac{a}{2}+r_{c}\right)^{2}=r_{c}^{2}+\left(p-\frac{a}{2}\right)^{2}$$

Then $a r_{c}=p(p-a)$. Since $r_{c}=\frac{S}{p-c}$, we obtain by using Heron's formula
$$a S=p(p-a)(p-c)=\frac{S^{2}}{p-b}$$
$$a(p-b)=S$$
Since $a$, $b$ and $c$ form (in this order) an arithmetic progression, we have $a=b-x$, $c=b+x$ and
$$p=\frac{3b}{2},\quad p-a=\frac{b}{2}+x,\quad p-b=\frac{b}{2},\quad p-c=\frac{b}{2}-x$$
Now by (1) and the Heron formula we obtain the equation
$$(b-x)^{2}=3\left(\frac{b^{2}}{4}-x^{2}\right)$$
which has a unique solution $x=\frac{b}{4}$. Therefore $a=\frac{3b}{4}$, $c=\frac{5b}{4}$ and then $a^{2}+b^{2}=c^{2}$, i.e. $\text{ACB=90}$.