Square is Sum of Two Rectangles

Proof

 * Euclid-II-2.png

Let $AB$ be the given straight line cut at random at the point $C$.

Construct the square $ABED$ on $AB$.

Construct $CF$ parallel to $AD$.

Then $\Box ABED = \Box ACFD + \Box CBEF$.

Now $\Box ABED$ is the square on $AB$.

Similarly, from Opposite Sides and Angles of Parallelogram are Equal:
 * $\Box ACDF$ is the rectangle contained by $AB$ and $AC$, as $AB = AD$
 * $\Box CBFE$ is the rectangle contained by $AB$ and $BC$, as $AB = AD$.

Hence the result.