smc

For the sake of clarity, let the outermost triangle be $ABC$ with incircle tangency points $D$, $E$, and $F$ on $BC$, $AC$ and $AB$ respectively. Let $\angle A=\alpha$, and $\angle B=\beta$. Because $△AFE$ and $△BDF$ are isosceles, $\angle AFE=\frac{180-\alpha }{2}$ and $\angle BFD=\frac{180-\beta }{2}$. So $\angle DFE=180-\frac{180-\alpha }{2}-\frac{180-\beta }{2}=\frac{\alpha +\beta }{2}$, and since $\alpha +\beta <180^{\circ }$, $\angle DFE$ is acute. The same method applies to $\angle FED$ and $\angle FDE$, which means $△DEF$ is acute - hence our answer is $\boxed{\text{are always acute angles}}$.