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

 * 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.


 * 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.