Tintarn wrote: Let $ABC$ be a triangle with incenter $I$. Let $M_b$ and $M_a$ be the midpoints of $AC$ and $BC$, respectively. Let $B'$ be the point of intersection of lines $M_bI$ and $BC$, and let $A'$ be the point of intersection of lines $M_aI$ and $AC$. If triangles $ABC$ and $A'B'C$ have the same area, what are the possible values of $\angle ACB$? Using barycentric coordinates we get that the area condition is equivalent to $a^2-ab+b^2=c^2$. Using the law of cosines this is eqiuvalent to $\cos\gamma=\frac{1}{2}$ and $\gamma=60^\circ$.

We use barycentric coordinates: $A = (1:0:0), \ B = (0:1:0), \ M_a = (0:1:1), \ M_b = (1:0:1)$. Let $CA' = t$, then $A'A = b-t$, therefore $A' = (t:0:b-t)$. Now, $I = (a:b:c)$ is the incentreand we have that $M_a, I, A'$ are collinear, therefore,$$ \begin{vmatrix} a & b &c\\ 0 & 1 & 1\\ t & 0 & b-t \end{vmatrix} = 0.$$Thus, $t(b-c) + (b-t)(a) = 0$. So, we get that $CA' = \frac{ab}{a+c-b}$. Similarly, $CB' = \frac{ab}{a+b-c}$. Now, $\frac 12 ab \sin C = [ABC] = [AB'C'] = \frac 12 CA' \cdot CB' \sin C$, therefore, $ab = (a+c-b)(a+b-c)$. This simplifies to $a^2-ab+b^2 = c^2$, which as noted above implies that $\cos C = \frac 12$ by the cosine rule, and therefore $\angle ACB = C = 60^{\circ}$.

I love this problem. Wondered whether there is a 4-piece dissection of the triangles ABC, A'B'C (see Greg N. Frederickson, "Dissections: Plane & Fancy", Cambridge University Press, 1997)? I only found a 5-piece dissection (not yet allowed to post the respective image, unfortunately).