Square of Sum
Contents |
Theorem
- $\forall x, y \in \R: \left({x + y}\right)^2 = x^2 + 2 x y + y^2$
Algebraic Proof
Follows from the distribution of multiplication over addition:
- $(x + y)^2 = (x + y) \cdot (x + y) = x \cdot (x + y) + y \cdot (x + y) = x \cdot x + x \cdot y + y \cdot x + y \cdot y = x^2 + 2xy + y^2$
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 $n = 2$.
Geometric Proof
As Euclid put it:
- "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 Parallel Implies Equal Alternate Interior Angles, $\angle CGB = \angle ADB$.
We have that $BA = AD$.
So from Isosceles Triangles have Two Equal Angles, $\angle ADB = \angle ABD$.
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 $CG \| BK$, from Parallel 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.
$\blacksquare$
Historical Note
This is Proposition 4 of Book II of Euclid's The Elements.
