Under what conditions do the angle bisectors of a triangle make $120^\circ$ angles at the incenter?

Question

I haven't messed around with geometry in a while.

What criteria does this triangle need to meet so you can state that the incircle angles $\alpha,\beta, \gamma $ here equal 120º? Do any of the sides have to have same length? Right angles? Anything?

I tried using the law of cosines, but that doesn't seem very fruitful. Thanks in advance.

Answer 1

Since $\alpha +A/2 + C/2 = 180$, it follows that $A/2+C/2=60$, or $A+C=120$. Thus, $B=60$. Similarly, $A=60$ and $B=60$. So $\triangle ABC$ is equilateral.