Solution:

First, a lemma: If squares constructed on $X Y$ and $Y Z$ of triangle $X Y Z$ have respective centers $U, V$, then a rotation through angle $\pi / 2$ about the midpoint $M$ of $Z X$ carries $U$ into $V$.

The proof is as follows: Rotating quadrilateral $M X U Y$ by an angle $\pi$ about $M$ gives us a quadrilateral $M Z W T$ (we know that $Z$ is the image of $X$ because $M$ is the midpoint of $Z X$). Now, a rotation by angle $\pi / 2$ about $V$ carries $Z$ into $Y$. We will show that this same rotation carries $W$ into $U$. For this it suffices to check that triangles $V Z W$, $V Y U$ are congruent (and equally oriented). Since $V Z = V Y$, $Z W = X U = Y U$, and (using directed angles if necessary)
$$\angle W Z V = \pi / 2 + \angle T Z Y = 3 \pi / 2 - \angle Z Y X = \angle U Y V$$
the congruence holds. So, as claimed, a $\pi / 2$ rotation about $V$ takes $W$ into $U$, so that $\triangle U V W$ is an isosceles right triangle, and $M$ is the midpoint of its hypotenuse. The lemma then follows.

In our original situation now, let the squares on $A B$, $B C$, $C D$, $D A$ have respective centers $E, F, G, H$. By the lemma, a rotation of angle $\pi / 2$ about the midpoint of $A C$ maps $E$ into $F$. This rotation also maps $G$ into $H$. Hence it takes the segment $E G$ into $F H$, so these segments are equal and perpendicular.