Sum of Squares of Sum and Difference/Geometric Proof 2

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 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.


 * Euclid-II-10.png

That is, from the above diagram: $\left({AC + CD}\right)^2 + \left({CD - AC}\right)^2 = 2 \left({AC^2 + CD^2}\right)$.

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 Interior 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.