Rectangle is Sum of Square and Rectangle

Theorem

In the words of Euclid:

If a straight line be cut at random, the rectangle contained by the whole and one of the segments is equal to the rectangle contained by the segments and the square on the aforesaid segment.

(The Elements: Book $\text{II}$: Proposition $3$)

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 proof is Proposition $3$ of Book $\text{II}$ of Euclid's The Elements.
This is little more than an example of Proposition $1$: Real Multiplication Distributes over Addition, and could be directly derived from it. Euclid, for some reason, preferred not to do this.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{II}$. Propositions