Solution:

Set $\angle CAB = \alpha$, $\angle ABC = \beta$ and $\angle BCA = \gamma$. Then
$$\text{IPC 1 = 1 2 ( BA 1 + C 1B 1 ) = 1 2 ( BA 1 + AC 1 + AB 1 ) = 1 2 ( + + ) = 90 .}$$
Hence $O$ is the midpoint of the segment $IC_1$ and it follows that

$$\text{IOP = 2 IC 1P = CA 1 = .}$$
We also have $\angle CC_1B = \alpha$ and therefore $OP \parallel C_1B$. Since $C_1O = OI$ and $BM = MN$, we conclude that $IN \parallel C_1B$, i.e. $\angle CIA_1 = \alpha$. On the other hand,
$$\text{CIA 1 = 1 2 ( CA 1 + AC 1 ) = 1 2 ( + ) = 90 - 2 .}$$
This implies that $\alpha = 90^\circ - \frac{\beta}{2} = \frac{\alpha + \gamma}{2}$, i.e. $\alpha = \gamma$. Since $\alpha = 2\beta$, we obtain that $\alpha = \gamma = 72^\circ$ and $\beta = 36^\circ$.