In a triangle ABC with usual notations if, $\cot \frac{A}{2}=\frac{b+c}{a}$, then the triangle $A B C$ is

Given the condition in triangle ABC: $\cot \frac{A}{2} = \frac{b + c}{a}$.

Using the Law of Sines, $a = 2R \sin A$, $b = 2R \sin B$, $c = 2R \sin C$, substitute into the equation:

$\cot \frac{A}{2} = \frac{\sin B + \sin C}{\sin A}$.

Apply sum-to-product and double-angle identities:

$\sin B + \sin C = 2 \sin \left(\frac{B + C}{2}\right) \cos \left(\frac{B - C}{2}\right)$
and $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$.

From the angle sum $A + B + C = \pi$, note that $\frac{B + C}{2} = \frac{\pi - A}{2} = \frac{\pi}{2} - \frac{A}{2}$,
so $\sin \left(\frac{B + C}{2}\right) = \cos \frac{A}{2}$.

Substitute and simplify:

$\cot \frac{A}{2} = \frac{2 \cos \frac{A}{2} \cos \left(\frac{B - C}{2}\right)}{2 \sin \frac{A}{2} \cos \frac{A}{2}} = \frac{\cos \left(\frac{B - C}{2}\right)}{\sin \frac{A}{2}}$.

Since $\cot \frac{A}{2} = \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}$, equate:

$\cos \frac{A}{2} = \cos \left(\frac{B - C}{2}\right)$.

Given the domain constraints, $\frac{A}{2} = \pm \frac{B - C}{2}$, leading to $A = B - C$ or $A = C - B$.

Substitute into $A + B + C = \pi$:
For $A = B - C$, $B - C + B + C = \pi \implies B = \frac{\pi}{2}$.
For $A = C - B$, $C - B + B + C = \pi \implies C = \frac{\pi}{2}$.

In both cases, triangle ABC is right-angled.

The correct choice is $\boxed{C}$.