$ [AC'BA'CB'] = 4[ABU]+4[BCU]+4[CAU]-2S=4S-2S=2S$, as desired.

robbo wrote: $ [AC'BA'CB'] = 4[ABU]+4[BCU]+4[CAU]-2S=4S-2S=2S$, as desired. our Dear robbo Maybe attached drawing could help, to better understand your solution. I chose point U as the intersection of median lines of triangle. Nice solution.

Attachments:

Shu wrote: Let $ABC$ denote a triangle with area $S$. Let $U$ be any point inside the triangle whose vertices are the midpoints of the sides of triangle $ABC$. Let $A'$, $B'$, $C'$ denote the reflections of $A$, $B$, $C$, respectively, about the point $U$. Prove that hexagon $AC'BA'CB'$ has area $2S$. What do you mean by "whose vertices" in a point? Do you mean U is the centeriod of the triangle? For some arbitrary points U inside the triangle, there is no proper polygon, but also it is not necessary that U be the centeriod of the triangle. the location of U should be limited.

To explain the question; the point $U$ is in that triangle, because the reflections would be outside the triangle, to get a convex hexagon. It is then easy to see that $[CUBA']=[CUBA]$ and similer,adding these, we get $4S$ and with substracting $2S$ for sum of $[CUB].$ (so adding 3 quadrilaterals with area twice the traingle and substracting the 3 triangles)

yabi wrote: What do you mean by "whose vertices" in a point? Do you mean U is the centeriod of the triangle? For some arbitrary points U inside the triangle, there is no proper polygon, but also it is not necessary that U be the centeriod of the triangle. the location of U should be limited. I apologize if this wasn't clear. To clarify, the point $U$ can be any point inside the medial triangle.