Two side-by-side squares are inscribed in a semicircle. The diameter of the semicircle is 16. What is the sum of the two squares' areas?

A somewhat devious way is to extract the fact that the answer doesn't depend on where D is, so place it at the origin!

Then it's simply:

$$ x = y $$

$$ \therefore x^2 + y^2 = 2 x^2 = 8^2 $$

Let $OD=c$ and the side lengths of the two squares $a$ and $b$. From right triangles OAB and OHF,

$$r^2=a^2+(a-c)^2;\>\>\>\>\> r^2=b^2+(b+c)^2$$

Eliminate $c$ to get,

$$a-\sqrt{r^2-a^2}=\sqrt{r^2-b^2}-b$$

Square both sides,

$$a\sqrt{ r^2-a^2} =b\sqrt{r^2-b^2}$$

Square again and rearrange,

$$r^2(a^2-b^2)=a^4-b^4$$

Thus, the sum of the two areas is

$$a^2+b^2=r^2=64$$

Let $V$ be the common vertex of our squares which lies on the diameter of the semicircle. We may assume that the coordinates of $V$ are $(x,0)$. The upper-right-corner $A$ is located at the intersection (the one with a positive ordinate) of $x^2+y^2=64$ and $y=x-v$. The upper-left-corner $B$ is located at the intersection (always the one with a positive ordinate) of $y=v-x$ and $x^2+y^2=64$. It follows that the ordinate of $A$ is $\frac{1}{2}\left(-v+\sqrt{128-v^2}\right)$ and the ordinate of $B$ is $\frac{1}{2}\left(v+\sqrt{128-v^2}\right)$. By summing the squares of these numbers we get that the total area of our squares is $$ \frac{1}{4}(v^2+128-v^2+v^2+128-v^2) = 64,$$ i.e. the area of a square built on a radius.

In order to produce an elementary proof, we just have to show that the length of $AB=\sqrt{2}\sqrt{AA'^2+BB'^2}$ does not depend on the position of $V$ on the diameter. But this is trivial since the symmetric of $B$ with respect to the diameter "sees" the $AB$-chord under an angle of $45^\circ$, such that $\widehat{AOB}$ always is a right angle (and $AOVB$ is a cyclic quadrilateral).