# Square of Sum

## Theorem

$\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$

## Algebraic Proof 1

Follows from the distribution of multiplication over addition:

 $\ds \paren {x + y}^2$ $=$ $\ds \paren {x + y} \cdot \paren {x + y}$ Definition of Square Function $\ds$ $=$ $\ds x \cdot \paren {x + y} + y \cdot \paren {x + y}$ Real Multiplication Distributes over Addition $\ds$ $=$ $\ds x \cdot x + x \cdot y + y \cdot x + y \cdot y$ Real Multiplication Distributes over Addition $\ds$ $=$ $\ds x \cdot x + x \cdot y + x \cdot y + y \cdot y$ Real Multiplication is Commutative $\ds$ $=$ $\ds x \cdot x + 2 x y + y \cdot y$ simplifying $\ds$ $=$ $\ds x^2 + 2 x y + y^2$ Definition of Square Function

$\blacksquare$

## Algebraic Proof 2

Follows directly from the Binomial Theorem:

$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$

putting $n = 2$.

$\blacksquare$

## Geometric Proof

In the words of Euclid:

If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments. Let the straight line $AB$ be cut at random at $C$.

Construct the square $ADEB$ on $AB$ and join $DB$.

Construct $CF$ parallel to $AD$ through $C$.

Construct $HK$ parallel to $AB$ through $G$.

From Parallelism implies Equal Alternate Angles, $\angle CGB = \angle ADB$.

We have that $BA = AD$.

$\angle ADB = \angle ABD$

So $\angle CGB = \angle CBG$, and so from Triangle with Two Equal Angles is Isosceles, $BC = CG$.

From Opposite Sides and Angles of Parallelogram are Equal, we have $CB = GK$ and $CG = KB$ and so $CGKB$ is equilateral.

Now, since $CG \parallel BK$, from Parallelism implies Supplementary Interior Angles we have that $\angle KBC$ and $\angle GCB$ are supplementary.

But as $\angle KBC$ is a right angle, $\angle BCG$ is also a right angle.

So from Opposite Sides and Angles of Parallelogram are Equal, $\angle CGK$ and $\angle GKB$ are also right angles.

So $CGKB$, described on $CB$. is right-angled, and as it is equilateral, by definition it is square.

For the same reason, $HGDF$ is also square, and described on $HG$, which equals $AB$.

So $HGDF$ and $CGKB$ are the squares on $AC$ and $CB$.

Now we have from Complements of Parallelograms are Equal that $\Box ACGH = \Box GKEF$.

But $\Box ACGH$ is the rectangle contained by $AC$ and $CB$, as $CB = GC$.

So $\Box GKEF$ is also equal to the rectangle contained by $AC$ and $CB$.

But the squares $HGFD$ and $CBKG$ are equal to the squares on $AC$ and $CB$.

So the four areas $HGFD$, $CBKG$, $ACGH$ and $GKEF$ are equal to:

the squares on $AC$ and $CB$

and:

twice the rectangle contained by $AC$ and $CB$.

But $HGFD$, $CBKG$, $ACGH$ and $GKEF$ are also equal to the square on $AB$.

Hence the result.

$\blacksquare$