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

Proof

 * Euclid-XIII-3.png

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