Sum of Squares of Sum and Difference/Geometric Proof 1

Theorem

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

Proof

 * Euclid-II-9.png

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.