# Area of Square on Greater 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 greater segment added to the half of the whole is five times the square on the half.*

(*The Elements*: Book $\text{XIII}$: Proposition $1$)

## Proof

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

Let $AC$ be the greater segment.

Let the straight line $AD$ be produced in a straight line with $CA$.

Let $AD = \dfrac {AB} 2$.

It is to be demonstrated that:

- $CD^2 = 5 \cdot AD^2$

Let the squares $AE$ and $DF$ be drawn on $AB$ and $DC$.

Let the figure in $DF$ be drawn.

Let $FC$ be produced to $G$.

We have that $AB$ has been cut in extreme and mean ratio at $C$.

Therefore from:

and:

it follows that:

- $AB \cdot BC = AC^2$

From the construction:

- $CE = AB \cdot BC$
- $FH = AC^2$

Therefore:

- $CE = FH$

We have that:

- $BA = 2 \cdot AD$

while:

- $BA = KA$
- $AD = AH$

Therefore:

- $KA = 2 \cdot AH$

But from Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base

- $KA : AH = CK : CH$

Therefore:

- $CK = 2 \cdot CH$

But:

- $LH + HC = 2 \cdot CH$

Therefore:

- $KC = LH + HC$

But:

- $CE = HF$

Therefore $AE$ equals the gnomon $MNO$.

We have that:

- $BA = 2 \cdot AD$

Therefore:

- $BA^2 = 4 \cdot A^2$

That is:

- $AE = 4 \cdot DH$

But:

- $AE = MNO$

Therefore:

- $MNO = 4 \cdot AP$

Therefore:

- $DF = 5 \cdot AP$

But:

- $DF$ is the square on $DC$

and:

- $AP$ is the square on $DA$.

Hence the result.

$\blacksquare$

## Historical Note

This proof is Proposition $1$ of Book $\text{XIII}$ of Euclid's *The Elements*.

It is the converse of Proposition $2$: Converse of Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{XIII}$. Propositions