Prove that the constructed shape is a square We construct four squares outside a parallelogram on it's sides Prove that if we connect the centers of the squares we get a square.
My attempt:I connected the intersection of the parallelogram's diagonals to the centers by symmetry we can easily prove the shape is a parallelogram.But how to prove it is a square?

 A: $A,B,C,D$ are centers of squares on the side of a parallelogram.  Prove figure $ABCD$ is a square.  Join $AE$, $AF$, $EB$, $FD$. Triangles $AEB$ and $AFD$ are congruent, since $AE = AF$, $EB = FD$, and $\angle AEB = \angle AFD$, each angle being a right angle ($45+45$) plus the supplement of the angle of the parallelogram at $F$. Therefore, $AB = AD$. And since $\angle EAB = \angle FAD$, then by addition $\angle EAF = \angle BAD$. But $\angle EAF$ is right. Therefore $\angle BAD$ is right. In this way we prove $ABCD$ is equilateral and right-angled. 
A: Let $z$, $w$, $-z$, $-w\in{\mathbb C}$ be the vertices of the parallelogram in counterclockwise order. The square over the edge $[z,w]$ then has vertices $v_0=z$, $v_1=z-i(w-z)$, $v_2=w+i(z-w)$, $v_3=w$ and center $$c={1\over4}(v_0+v_1+v_2+v_3)={1-i\over 2}(w+iz)\ .$$
The replacement $z\rightsquigarrow w$, $w\rightsquigarrow -z$ produces the next center
$$c'={1-i\over2}\bigl(-z+iw)=i c\ ,$$
and it becomes obvious that $c''=ic'$, $c'''=i c''$. All of this shows that the points $c$, $c'$, $c''$, $c'''$ form the vertices of a square.
