Let $ABC$ be a triangle with incenter $I$, midpoints $M_a$ and $M_b$ of sides $a$ and $b$ respectively. Let $A'$ be the intersection of $M_aI$ and $AC$ and $B'$ the intersection of $M_bI$ and $BC$. The area of $ABC$ and $A'B'C$ is equal.
Every equilateral triangle satisfies these conditions since $A=A'$ and $B=B'$. This can be simply proven by congruence of $\Delta ABM_b$ and $\Delta CBM_b$. Vice versa for $A$: $\Delta ABM_a \cong \Delta CAM_a$. This implies that $AM_a$ resp. $BM_b$ are angle bisectors, so $A,I,M_a$ resp. $B,I,M_b$ are collinear. So $A=A'$ and $B=B'$.
Are there any other triangles that satisfy these conditions?
Edit: Some diagrams for non-equilateral triangles and one example of another triangle that works:
If $\angle ACB = \alpha$ is small and $ABC$ is isosceles or $AC$ and $BC$ are roughly the same length, $A'$ and $B'$ seem to lie outside $ABC$.
If $\alpha$ is bigger, $A'$ and $B'$ seem to lie inside of $ABC$
The only cases for which the two areas may be equal seem to bo if $AC$ and $BC$ are of notable different length. Then $AB$ and $A'B'$ seem to intersect:
A 30-60-90-triangle seems to work too:
But I don't know if there exist suitable $ABC$ for $\alpha \neq 60$