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

Let $AC$ be bisected at $D$.

It is to be demonstrated that $BD^2 = 5 \cdot DC^2$.

Let the squares $AE$ be described on $AB$.

Let the figure be drawn as above.

We have that:

$AC = 2 \cdot DC$

Therefore:

$AC^2 = 4 \cdot DC^2$

That is:

$RS = 4 \cdot FG$

We have by definition of extreme and mean ratio that:

$AB \cdot BC = AC^2$

and:

$CE = AB \cdot BC$

Therefore:

$CE = RS$

But:

$RS = 4 \cdot FG$

Therefore:

$CE = 4 \cdot FG$

We have that:

$AD = DC$

and so:

$HK = KF$

Hence:

$GF = HL$

Therefore:

$GK = KL$

that is:

$MN = NE$

Hence:

$MF = CG$

Therefore:

$CG = FE$

and so:

$CG + CN = FE + CN$

That is, the gnomon $OPQ$ equals the rectangle $CE$.

But $CE = 4 \cdot GF$

Therefore:

$OPQ = 4 \cdot GF$

Therefore:

$OPQ + FG = 5 \cdot GF$

But:

$OPQ + FG = DN$

while:

$DN = DB^2$

and:

$GF = DC^2$

Therefore:

$BD^2 = 5 \cdot DC^2$

$\blacksquare$

## Historical Note

This proof is Proposition $3$ of Book $\text{XIII}$ of Euclid's The Elements.