Rectangle is Sum of Square and Rectangle

Proof

 * Euclid-II-3.png

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

Construct the square $CDEB$ on $AB$.

Produce $ED$ to $F$ and construct $AF$ parallel to $CD$.

Then $\Box ABEF = \Box ACDF + \Box CBED$.

Now $\Box ABEF$ is the rectangle contained by $AB$ and $BC$, as $BC = CD = AF$.

Similarly, from Opposite Sides and Angles of Parallelogram are Equal:
 * $\Box ACDF$ is the rectangle contained by $AC$ and $BC$, as $BC = AF$
 * $\Box CBED$ is the square on $BC$.

So the rectangle contained by $AB$ and $BC$ equals the rectangle contained by $AC$ and $BC$ together with the square on $BC$.