Area of Squares on Whole and Lesser Segment of Straight Line cut in Extreme and Mean Ratio

Proof

 * Euclid-XIII-4.png

Let the line $AB$ be cut in extreme and mean ratio at the point $C$.

Let $AC$ be the greater segment.

It is to be demonstrated that:
 * $AB^2 + BC^2 = 3 \cdot CA^2$

Let the square $ADEB$ be described on $AB$.

Let the figure be drawn as above.

We have that $AB$ be cut in extreme and mean ratio at $C$ such that $AC > CB$.

Therefore from:

and:

it follows that:
 * $AB \cdot BC = AC^2$

We have that:
 * $AK = AB \cdot BC$

and:
 * $HG = AC^2$

Therefore:
 * $AK = HG$

We have that:
 * $AF = FE$

So:
 * $AF + CK = FE + CK$

Therefore:
 * $AK = CE$

and so:
 * $AK + CE = 2 \cdot CE$

But:
 * $AK + CE$ are the gnomon $LMN$ and the square $CK$.

Therefore:
 * $LMN + CK = 2 \cdot AK$

But we have that:
 * $AK = HG$

Therefore:
 * $LMN + CK + HG = 3 \cdot HG$

But:
 * $LMN + CK + HG = AE + CK$

while:
 * $HG = AC^2$

Therefore:
 * $AB^2 + BC^2 = 3 \cdot CA^2$