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

## Theorem

In the words of Euclid:

If a straight line be cut in extreme and mean ratio, the square on the whole and the square on the lesser segment together are triple of the square on the greater segment.

## Proof 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:

Proposition $17$ of Book $\text{VI}$: Rectangles Contained by Three Proportional Straight Lines

and:

Book $\text{VI}$ Definition $3$: Extreme and Mean Ratio

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:

$AB^2 + BC^2 = 3 \cdot CA^2$

$\blacksquare$

## Historical Note

This proof is Proposition $4$ of Book $\text{XIII}$ of Euclid's The Elements.