Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio

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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

Euclid-XIII-1.png

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:

$KA = 2 \cdot AH$

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 theorem 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