Square of Sum with Double

Theorem

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

Algebraic Proof
Follows from the distribution of multiplication over addition:
 * $(a + 2b)^2 = (a + 2b)\cdot(a + 2b) = a \cdot (a + 2b) + 2b \cdot (a + 2b) = a \cdot a + a \cdot 2b + 2b\cdot a + 2b \cdot 2b = a^2 + 4ab + 4b^2$

More succinctly, it follows directly from the Binomial Theorem:
 * $\forall n \in \Z_+: \left({x+y}\right)^n = \sum_{k=0}^n {n\choose k}x^{n-k}y^k$

putting $n = 2, x = a, y = 2b$.

Geometric Proof
As Euclid put it:


 * "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: $4 \left({a + b}\right) b + a^2 = \left({a + 2b}\right)^2$.


 * 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 $ADFE$ 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, we have:
 * $\Box CBGK = \Box BDNG$;
 * $\Box KGRO = \Box GNPR$.

Also, from Complements of Parallelograms are Equal, we have that $\Box CBGK = \Box GNPR$, and so $\Box BDNG = \Box KGRO$.

So all of them are equal: $\Box CBGK = \Box BDNG = \Box KGRO = \Box GNPR$.

So the four of them together are four times $\Box CBGK$.

Similarly, by Parallelograms with Equal Base and Same Height have Equal Area, we have:
 * $\Box ACKM = \Box MKGQ$;
 * $\Box GRHL = \Box RPFL$.

Also, from Complements of Parallelograms are Equal, we have that $\Box MKGQ = \Box GRHL$, and so $\Box ACKM = \Box RPFL$.

So all of them are equal: $\Box ACKM = \Box MKGQ = \Box GRHL = \Box RPFL$.

So the four of them together are four times $\Box ACKM$.

Adding all these eight areas together, we see that the gnomon $ADFHOQ$ equals four times $\Box CBGK$ plus four times $\Box ACKM$, or four times $\Box ABGM$.

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

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

Now we add $\Box QOHE$ to each. Note that $\Box QOHE$ equals the square on $AC$.

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

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

Hence the result.