Square of Sum with Double/Geometric Proof

Theorem

 * $\forall a, b \in \R: \left({a + 2 b}\right)^2 = a^2 + 4 a b + 4b^2$

Proof
That is: $4 \left({a + b}\right) b + a^2 = \left({a + 2 b}\right)^2$.


 * Euclid-II-8.png

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

Then four times the rectangle contained by $AB$ and $BC$ together with the square on $AC$ equals the square on $AB$ and $BC$ as a single straight line.

The proof is as follows.

Produce $AB$ to $D$ where $BD = BC$.

Construct the square $AEFD$ on $AD$ and join $DE$.

Draw the given figure above, in the same manner as in Square of Sum.

By Parallelograms with Equal Base and Same Height have Equal Area:
 * $\Box CBKG = \Box BDNK$
 * $\Box KGQR = \Box KNPR$

From Complements of Parallelograms are Equal:
 * $\Box CBKG = \Box KNPR$

and so:
 * $\Box BDNK = \Box KGQR$

So all of them are equal:
 * $\Box CBKG = \Box BDNK = \Box KGQR = \Box KNPR$

So the four of them together equal $4 \Box CBKG$.

Similarly, by Parallelograms with Equal Base and Same Height have Equal Area:
 * $\Box ACGM = \Box MGQO$
 * $\Box QRLH = \Box RPFL$

From Complements of Parallelograms are Equal:
 * $\Box MGQO = \Box QRLH$

and:
 * $\Box ACGM = \Box RPFL$

So all of them are equal:
 * $\Box ACGM = \Box MGQO = \Box QRLH = \Box RPFL$

So the four of them together equal $4 \Box ACGM$.

Adding all these eight areas together, we see that the gnomon $STU$ equals $4 \Box CBKG + 4 \Box ACGM = 4 \Box ABKM$.

But $\Box ABKM$ is the rectangle contained by $AB$ and $BC$, as $BC = BK$.

So four times the rectangle contained by $AB$ and $BC$ equals the area of the gnomon $STU$.

Now we add $\Box QOEH$ to each.

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

So four times the rectangle contained by $AB$ and $BC$, together with the square on $AC$, equals the gnomon $STU$ together with $\Box QOEH$.

But the gnomon $STU$ together with $\Box QOEH$ form the whole square $ADFE$, which is the square on $AB$ and $BC$ as a single straight line.

Hence the result.