Three isosceles triangles △ABC', △AB'C, and △A'BC were constructed on the sides of triangle ABC. Three isosceles triangles $△ABC'$, $△AB'C$, and $△A'BC$ were constructed on the sides of triangle $ABC$. The vertex angles have the following measurements; $∠A'=α, ∠B'=β, ∠C'=γ$. Prove that if $α+β+γ=360°$ then the measurements of the angles of $△A'B'C'$ do not depend on the measurements of the angles of $△ABC$.

$D$, $E$, and $F$ are the midpoints of the sides of $△ABC$, making $A'E$, $B'F$, and $C'D$ the perpendicular bisectors of their respective triangles.
Let us consider the circles $\Gamma_A,\Gamma_B,\Gamma_C$ centered at $A',B',C'$ with radii $A'B,B'C,C'A$.
The constraint on the angles ensures that these circles meet at a point $P$.

We also have that $PB'AC', PC'BA'$ and $PA'CB'$ are kites, therefore $\widehat{B'A' C'}=\frac{1}{2}\alpha$ and so on.
Comment: These figures helps you the solution: In figure (b) the triangle is equilateral and $A'=B'=C'=120^o$ so angles of triangle A'B'C' are all $60^o=\frac{A'}2=\frac{B'}2=\frac{C'}2$
In figure (a) The sum of interior angle of triangle A'B'C' is $180^o=\frac{\alpha +\beta+\gamma}2$. Noting the fact that every triangle can be considered as a transformed equilateral triangle, keeping the point O which is intersection of perpendicular bisectors fixed, one may conclude that angles in triangle A'B'C' keep their particularities and have these values: $B'A'C'=\frac{\alpha}2$, $A'B'C'=\frac{\beta}2$ and $B'C'A'=\frac{\gamma}2$ which is not a function of the angles of triangle ABC.
PS: Figure d is transformed of figure d. Point O , measure of side AC and its distance to O is kept constant. Vertex B is moved to be co-linear with perpendicular bisector of AC. In all positions of B the conditions hold including when AB=BC=AC which is shown in figure d.Reverse action converts figure d to c.