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

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:

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
 * $KA : AH = CK : CH$

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.