Rectangle is Sum of Square and Rectangle

Theorem

If a straight line is cut at random, the rectangle contained by the whole and one of the segments equals rectangle contained by the segments and the square on the aforementioned segment.

Proof

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 $BC$.

$\blacksquare$

Historical Note

This is Proposition 3 of Book II of Euclid's The Elements.

This is little more than an example of Book II Proposition 1, and could be directly derived from it. Euclid, for some reason, preferred not to do this.