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

 * If a straight line be cut at random, four times the rectangle contained by the whole and one of the segments together with the square on the remaining segment is equal to the square on the whole and the aforesaid segment as on one straight line.

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.