Square is Sum of Two Rectangles
Jump to navigation
Jump to search
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
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
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{II}$. Propositions