Label the squares' side lengths $a, b, c, d, e, f $ (clockwise from $A$). The claim is that $$a^2+c^2+e^2=b^2+d^2+f^2$$
Let $x$ be the altitude from $A$.
Let $y$ be the altitude from $B$.
Let $z$ be the altitude from $C$.
By the Pythagorean theorem applied to the two right triangles that include the altitude from $A$, we have:
$$x^2+c^2=(a+b)^2$$
$$x^2+d^2=(e+f)^2$$
By the Pythagorean theorem applied to the two right triangles that include the altitude from $B$, we have:
$$y^2+a^2=(e+f)^2$$
$$y^2+b^2=(c+d)^2$$
By the Pythagorean theorem applied to the two right triangles that include the altitude from $C$, we have:
$$z^2+e^2=(c+d)^2$$
$$z^2+f^2=(a+b)^2$$
Labeling the six Pythagorean equations above $(1)$ through $(6)$, we can add $(1)$, $(3)$, and $(5)$ to get:
$$ x^2+y^2+z^2 +a^2+c^2+e^2=(a+b)^2+ (c+d)^2 + (e+f)^2$$
Add $(2)$, $(4)$, and $(6)$:
$$ x^2+y^2+z^2 +b^2+d^2+f^2=(a+b)^2+ (c+d)^2 + (e+f)^2$$
Notice that the right sides of the above two equations are equal, so we may equate the left sides:
$$ x^2+y^2+z^2+a^2+c^2+e^2= x^2+y^2+z^2+b^2+d^2+f^2 $$
Now subtract $x^2+y^2+z^2$ from both sides, and we are done.
$$a^2+c^2+e^2=b^2+d^2+f^2$$