Square of Difference/Geometric Proof

Theorem

 * $\forall x, y \in \R: \left({x - y}\right)^2 = x^2 - 2 x y + y^2$

Proof

 * Euclid-II-7.png

(That is: $x^2 + y^2 = \left({x - y}\right)^2 + 2 x y$.)

Let the straight line $AB$ be cut at random at $C$.

Construct the square $ADEB$ on $AB$ and join $DB$.

Construct $CN$ parallel to $AD$ through $C$ and let it cross $DB$ at $G$.

Construct $HF$ parallel to $AB$ through $G$.

From Complements of Parallelograms are Equal, $\Box ACGH = \Box FGNE$, so add $\Box CBFG$ to each.

So the whole of $\Box ABFH$ equals the whole of $\Box CBEN$.

So $2 \Box ABFH = \Box ABFH + \Box CBEN$.

But $\Box ABFH + \Box CBEN$ equals the area of the gnomon $KLM$ together with $\Box CBFG$.

So the gnomon $KLM$ together with $\Box CBFG$ equals $2 \Box ABGH$.

But twice the rectangle contained by $AB$ and $BC$ is also equal to $2 \Box ABFH$, as $BG = BC$.

So the gnomon $KLM$ together with $\Box CBFG$ equals twice the rectangle contained by $AB$ and $BC$.

Add $\Box DHGN$ to each.

Note that $\Box DHGN$ equals the square on $AC$.

Then the gnomon $KLM$ together with $\Box CBFG$ and $\Box DHGN$ equals the whole of $\Box ABED$ and $\Box CBFG$, which are the squares on $AB$ and $BC$.

Hence the result.