Square of Sum less Square/Geometric Proof

Theorem

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

Proof
Let $AB$ be bisected at $C$, and let $BD$ be added to it in a straight line.

Then the rectangle contained by $AD$ and $BD$ together with the square on $BC$ equals the square on $CD$.


 * Euclid-II-6.png

The proof is as follows.

Construct the square $CEFD$ on $CD$, and join $DE$.

Construct $BG$ parallel to $CE$ through $B$, and let $BG$ cross $DE$ at $H$.

Construct $KM$ parallel to $AD$ through $H$.

Construct $AK$ parallel to $DF$ through $A$.

As $AC = CB$, from Parallelograms with Equal Base and Same Height have Equal Area we have that $\Box ACLK = \Box CBHL$.

From Complements of Parallelograms are Equal, $\Box CBHL = \Box HMFG$.

So $\Box ACLK = \Box HMFG$.

Add $\Box CDML$ to each.

So the whole of $\Box ADMK$ is equal to the gnomon $CDFGHL$.

But $\Box ADMK$ is the rectangle contained by $AD$ and $BD$, because $BD = DM$.

So the gnomon $CDFGHL$ is equal in area to the rectangle contained by $AD$ and $BD$.

Now $\Box LHGE$ is equal to the square on $BC$.

Add $\Box LHGE$ to each of the gnomon $CDFGHL$ and $\Box ADMK$.

Then the gnomon $CDFGHL$ together with $\Box LHGE$ equals the rectangle contained by $AD$ and $BD$ and the square on $BC$.

But the gnomon $CDFGHL$ together with $\Box LHGE$ is the square $CDFE$.

Hence the result.