Square is Sum of Two Rectangles

Theorem

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

Proof

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.

$\blacksquare$

Historical Note

This is Proposition 2 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.