Let us denote successively $r$ and $r_a$ the radii of the inscribed circle and the exscribed circle wrt to side BC of triangle $ABC$. Prove that if it is true that $r+r_a=|BC|$ , then the triangle $ABC$ is a right one
Really great problem for storage
Let's denote $s-a=x; s-b=y; s-c=z$ to have easier notation.
From Heron's formula: $P^2=s(s-a)(s-b)(s-c)$ and $P=rs$ so $r^2=\frac{(s-a)(s-b)(s-c)}{s}=\frac{xyz}{s}$
Let $I_{a}$ be the center of exscribed triangle, so $P=P_{ABI_{a}}+P_{ACI_{a}}-P_{BCI_{a}}=r_{a}(\frac{b+c-a}{2})=r_{a}(s-a)$, so $r_{a}^2=\frac{(s(s-b)(s-c)}{s-a}=\frac{syz}{x}$. Now $r+r_{a}=a=(a+b+c)-b-c=2s-b-c=(s-b)+(s-c)=y+z$. Now squaring both sides we will get:
$r^2+2rr_{a}+r_{a}^2=y^2+2yz+z^2$ but $rr_{a}=\frac{P^2}{s(s-a)}$ and $zy=(s-b)(s-c)=\frac{P^2}{s(s-a)}$ so $rr_{a}=zy$ and now:
$r^2+r_{a}^2=y^2+z^2$ and from what we previously get: $\frac{xyz}{s}+\frac{syz}{x}=y^2+z^2$ or $x^2yz+s^2yz=z^2xs+y^2xs$. This can be represented as $(ys-zx)(zs-xy)=0$
We will put back and after long but easy factorization get:
$(ys-zx)(zs-xy)=(s(s-b)-(s-c)(s-a))(s(s-c)-(s-a)(s-b))=(a^2+c^2-b^2)(a^2+b^2-c^2)=0$
Now, since $(a^2+c^2-b^2)(a^2+b^2-c^2)=0$ it's easy to see that this triangle is right angled from Pythagoras.