Square of Sum less Square

Theorem
As Euclid put it:


 * "If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line."

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

Algebraic Proof
Follows directly and trivially from Square of Sum:

Geometric 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.