Sum of Squares of Sum and Difference/Geometric Proof 2

Proof

 * 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.