Given is a triangle $ABC$ with incircle center $I$ and side centers of $AC$ and $BC$ being named $M_a$ and $M_b$, respectively. Let the intersection of the lines $M_bI$ and $BC$ be called $B'$ and that of the lines $M_aI$ and $AC$ be named $A'$. Furthermore, the triangles $ABC$ and $A'B'C$ are equal in area.
How big can $\angle ACB$ be?
So far I have only come to the conclusion that $a' = k\cdot a \Rightarrow b' = \frac{b}{k}$ due to the equal areas and the common angle $\angle ACB$ by calculating the areas using $\frac{1}{2}ab\cdot \sin(\gamma) = A_1 = A_2 = \frac{1}{2}a'b'\cdot \sin(\gamma)$. Obviously that's close to nothing, so I really have no clue how to solve this task.
Edit: I have found out by trying that apparently the angle is always 60°, however, I still do not know how to prove that.
Let's look at the problem from a different angle. We may assume $C=(0,0)$, $A=(2,0)$, $B=(2s,2k)$, and $I=(x_I, y_I).$
So,
$$m_a=(s,k), \\ m_b=(1,0);$$
and it's very easy to show that:
$$B'=(\frac{sy_I}{sy_I-kx_I+k},\frac{ky_I}{sy_I-kx_I+k}), \\A'=(\frac{sy_I-kx_I}{y_I-k},0).$$
Now that we have the coordinates, we will obtain:
$$|CB'|=\frac{y_I \sqrt{s^2+k^2}}{sy_I-kx_I+k},\\ |CA'|=\frac{sy_I-kx_I}{y_I-k},\\ |CB|=2\sqrt{s^2+k^2}, \\|CA|=2, \\S_{\triangle A'B'C'}=\frac{1}{2} \times \sin C \times |CB'||CA'|=S_{\triangle ABC}=\frac{1}{2} \times \sin C \times |CB||CA| .$$
Therefore, we must have: $$\frac{y_I \sqrt{s^2+k^2}}{sy_I-kx_I+k}\times \frac{sy_I-kx_I}{y_I-k}=4\sqrt{s^2+k^2} \\ \implies y_I(sy_I-kx_I)=4(sy_I-kx_I+k)(y_I-k) \ (*).$$
On the other hand, we know that (take a look at this link):
$$(x_I, y_I)=(\frac{4\sqrt {s^2+k^2}+4s}{a+b+c}, \frac{4k}{a+b+c}),$$ where $c=|AB|=2\sqrt {(s-1)^2+k^2},$ $a=|BC|=2\sqrt {s^2+k^2}, $ and $b=|AC|=2.$
Thus, the identity $(*)$ yields the relations below: $$\frac{4k}{a+b+c}\times \frac{-4k\sqrt {s^2+k^2}}{a+b+c}=4\times (\frac {-4k\sqrt {s^2+k^2}}{a+b+c}+k)(\frac{4k}{a+b+c}-k) \\ \implies \frac{1}{a+b+c}\times \frac{-4\sqrt {s^2+k^2}}{a+b+c}=(\frac {-4\sqrt {s^2+k^2}}{a+b+c}+1)(\frac{4}{a+b+c}-1) \\ \implies -4\sqrt {s^2+k^2}=(-4\sqrt {s^2+k^2}+a+b+c)(4-a-b-c)\\ \implies -\sqrt {s^2+k^2}=(1-\sqrt {s^2+k^2}+\sqrt {(s-1)^2+k^2})(1-\sqrt {(s-1)^2+k^2}-\sqrt {s^2+k^2}) \\ \implies -\sqrt {s^2+k^2}=(1-\sqrt {s^2+k^2})^2-((s-1)^2+k^2) \\ \implies s^2-2s+1+k^2=1+s^2+k^2-\sqrt {s^2+k^2} \\ \implies 3s^2=k^2 \\ \implies \tan C=\frac{k}{s}=\sqrt 3 \implies C=60^{\circ}.$$
We are done.