# 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.

## 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:

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:

$AB \cdot BC = AC^2$

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:

$KA = 2 \cdot AH$
$KA : AH = CK : CH$

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.

$\blacksquare$