Similar triangle inside large triangle split by incenter

Consider an obtuse triangle $\bigtriangleup$ABC with $I$ being the incenter.

Among $\bigtriangleup$IAB, $\bigtriangleup$IBC and $\bigtriangleup$IAC, one of them is similar to $\bigtriangleup$ABC.

Find the ratio of the three angles of $\bigtriangleup$ABC.

Without loss of generality, let's suppose $\angle C$ is the obtuse angle.

Now, I suppose we should look at $\bigtriangleup$IAB, $\bigtriangleup$IBC and $\bigtriangleup$IAC one by one, for example, $\bigtriangleup$IAB can't be similar to $\bigtriangleup$CAB, given the fact that $\angle$C is clearly bigger than $\angle$AIB.

But how should we look at the other two triangles?

Solution 1:

Without lose of generality, suppose $\triangle IAC \sim \triangle CBA$. The similarity means congruency in the angles: $\angle ICA = \angle A$, $\angle CAI = \angle B$, and $\angle AIC = \angle C$.

Since the line from the vertex to incenter $\overline{AI}$ bisects $\angle A$, we have $\angle CAI = \frac12 \angle A$. The same reasoning ($\overline{CI}$ bisecting $\angle C$) also gives us $\angle ICA = \frac12 \angle C$.

Put the information together we have $$\left.\begin{aligned}&\angle B =\angle CAI = \frac12 \angle A \\ &\angle A= \angle ICA = \frac12 \angle C \end{aligned} \right\}\implies \angle A : \angle B : \angle C =2:1:4$$