Square is Sum of Two Rectangles

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Theorem

In the words of Euclid:

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

(The Elements: Book $\text{II}$: Proposition $2$)


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.

$\blacksquare$


Historical Note

This proof is Proposition $2$ 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