Square of Sum/Geometric Proof
Theorem
- $\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$
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.
(The Elements: Book $\text{II}$: Proposition $4$)
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 $BA = AD$.
So from Isosceles Triangle has 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 $CG \parallel BK$, 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.
As $CGKB$ is also 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 proof is Proposition $4$ of Book $\text{II}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{II}$. Propositions