Sum of Squares of Sum and Difference
Theorem
- $\forall a, b \in \R: \paren {a + b}^2 + \paren {a - b}^2 = 2 \paren {a^2 + b^2}$
Algebraic Proof
| \(\ds \) | \(\) | \(\ds \left({a + b}\right)^2 + \left({a - b}\right)^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 + 2 a b + b^2 + a^2 - 2 a b + b^2\) | Square of Sum and Square of Difference | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 + b^2 + a^2 + b^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \left({a^2 + b^2}\right)\) |
$\blacksquare$
Geometric Proof 1
In the words of Euclid:
- If a straight line be cut into equal and unequal segments, the squares on the unequal segments of the whole are double of the square on the half and of the square on the straight line between the points of section.
(The Elements: Book $\text{II}$: Proposition $9$)
That is, from the above diagram:
- $\left({AC + CD}\right)^2 + \left({AC - CD}\right)^2 = 2 \left({AC^2 + CD^2}\right)$
Let $AB$ be bisected at $C$, and let $D$ be another random point on $AB$.
Then the squares on $AD$ and $DB$ are double the squares on $AC$ and $CD$.
The proof is as follows.
Construct $CE$ perpendicular to $AB$ and construct $CE = AC$.
Join $EA$ and $EB$.
Construct $DF$ parallel to $CE$ to intersect $BE$ at $F$, and join $AF$.
Construct $FG$ parallel to $AB$ to intersect $CE$ at $G$.
Since $AC = EC$ we have from Isosceles Triangle has Two Equal Angles that $\angle EAC = \angle AEC$.
Since $\angle ACE$ is a right angle, from Sum of Angles of Triangle Equals Two Right Angles we have that $\angle EAC + \angle AEC$ also equals a right angle.
As they are equal, $\angle EAC$ and $\angle AEC$ are both equal to half a right angle.
For the same reason, $\angle EBC$ and $\angle BEC$ are also both equal to half a right angle.
We have that $\angle GEF = \angle BEC$, so $\angle GEF$ is also half a right angle.
From Parallelism implies Equal Corresponding Angles, $\angle EGF$ is a right angle.
So from Sum of Angles of Triangle Equals Two Right Angles, $\angle EFG$ is half a right angle.
So $\angle GEF = \angle GFE$, and so from Triangle with Two Equal Angles is Isosceles, $EG = GF$.
From Parallelism implies Equal Corresponding Angles, $\angle FDB$ is a right angle.
So from Sum of Angles of Triangle Equals Two Right Angles, $\angle BFD$ is half a right angle.
So $\angle BFD = \angle FBD$, and so from Triangle with Two Equal Angles is Isosceles, $FD = DB$.
Since $AC = CE$, the square on $AC$ equals the square on $CE$.
So the squares on $AC$ and $CE$ together are twice the square on $AC$.
But from Pythagoras's Theorem, the square on $AE$ equals the squares on $AC$ and $CE$ because $\angle ACE$ is a right angle.
So the square on $AE$ is double the square on $AC$.
By a similar argument, the square on $EF$ is double the square on $GF$.
But from Opposite Sides and Angles of Parallelogram are Equal, $GF = CD$.
So the square on $EF$ is double the square on $CD$.
So the squares on $AE$ and $EF$ are double the squares on $AC$ and $CD$.
We have that $\angle AEF$ is a right angle.
So from Pythagoras's Theorem, the square on $AF$ equals the squares on $AE$ and $EF$.
So the square on $AF$ is double the squares on $AC$ and $CD$.
We have that $\angle ADF$ is a right angle.
So from Pythagoras's Theorem, the square on $AF$ equals the squares on $AD$ and $DF$.
So the squares on $AD$ and $DF$ are double the squares on $AC$ and $CD$.
But $DF = DB$.
Hence the result.
$\blacksquare$
Geometric Proof 2
In the words of Euclid:
- If a straight line be bisected, and a straight line be added to it in a straight line, the square on the whole with the added straight line and the square on the added straight line both together are double of the square on the half and of the square described on the straight line made up of the half and the added straight line as on one straight line.
(The Elements: Book $\text{II}$: Proposition $10$)
That is, from the above diagram:
- $\paren {AC + CD}^2 + \paren {CD - AC}^2 = 2 \paren {AC^2 + CD^2}$
Let $AB$ be bisected at $C$, and let $AB$ be produced to some point $D$.
Then the squares on $AD$ and $BD$ are double the squares on $AC$ and $CD$.
The proof is as follows.
Construct $CE$ perpendicular to $AB$ and construct $CE = AC$.
Join $EA$ and $EB$.
Construct $EF$ parallel to $AB$, and Construct $DF$ parallel to $CE$.
From Parallelism implies Supplementary Interior Angles, $\angle CEF + \angle EFD$ equal two right angles.
Therefore $\angle FEB + \angle EFD$ is less than two right angles.
So by the Parallel Postulate, $EB$ and $FB$ must meet when produced in the direction $B$ and $D$.
Let this point be $G$, and let $AG$ be joined.
From Isosceles Triangle has Two Equal Angles, as $AC = AE$, $\angle EAC = \angle AEC$.
Because $\angle ACE$ is a right angle, from Sum of Angles of Triangle Equals Two Right Angles, each of $\angle EAC$ and $\angle AEC$ are each half a right angle.
For the same reason, each of $\angle CEB$ and $\angle EBC$ are each half a right angle.
Since $\angle EBC$ is half a right angle, from Two Straight Lines make Equal Opposite Angles, $\angle DBG$ is also half a right angle.
From Parallelism implies Equal Alternate Angles, $\angle BDG = \angle DCE$, so $\angle BDG$ is also a right angle.
So from Sum of Angles of Triangle Equals Two Right Angles, $\angle DGB$ is half a right angle.
As $\angle DGB = \angle DBG$, from Triangle with Two Equal Angles is Isosceles, we have that $BD = BG$.
From Opposite Sides and Angles of Parallelogram are Equal, $\angle EFD = \angle ECD$ which is a right angle,
Since $\angle EGF$ is half a right angle, and $\angle EFD$ is a right angle, $\angle FEG$ is half a right angle from Sum of Angles of Triangle Equals Two Right Angles.
So $\angle EGF = \angle FEG$, and so from Triangle with Two Equal Angles is Isosceles, $GF = EF$.
Since $EC = CA$, the square on $EC$ equals the square on $CA$.
So the squares on $EC$ and $CA$ are together twice the square on $CA$.
From Pythagoras's Theorem, the square on $EA$ is equal to the the squares on $EC$ and $CA$, and so twice the square on $AC$.
Again, since $FG = FE$, the square on $FG$ equals the square on $FE$.
So the squares on $FG$ and $EF$ are together twice the square on $FE$.
From Pythagoras's Theorem, the square on $EG$ is equal to the the squares on $GF$ and $FE$, and so twice the square on $EF$.
From Opposite Sides and Angles of Parallelogram are Equal, $EF = CD$.
But the square on $EA$ has been shown to be double the square on $AC$.
So the squares on $AE$ and $EG$ are twice the squares on $AC$ and $CD$.
From Pythagoras's Theorem, the square on $AG$ is equal to the the squares on $AE$ and $EG$.
So the square on $AG$is double the squares on $AC$ and $CD$.
But by Pythagoras's Theorem, the square on $AG$ is equal to the the squares on $AD$ and $DG$.
As $DG = DB$, the result follows.
$\blacksquare$