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

Proof

 * Euclid-XIII-2.png

Let the square on $AB$ be five times the square on $AC$.

Let $CD = 2 \cdot AC$.

It is to be demonstrated that when $CD$ is cut in extreme and mean ratio, the greater segment is $CB$.

Let the squares $AF$ and $CG$ be described on $AB$ and $CF$ respectively.

Let the figure $AF$ be drawn.

Let $BE$ be produced from $FB$.

We have that:
 * $BA^2 = 5 \cdot AC^2$

That is:
 * $AF = 5 \cdot AH$

Therefore the gnomon $MNO$ is $4$ times $AH$.

We have that:
 * $DC = 2 \cdot CA$

Therefore:
 * $DC^2 = 4 \cdot CA^2$

that is:
 * $CG = 4 \cdot AH$

But:
 * $MNO = 4 \cdot AH$

Therefore:
 * $MNO = CG$

We have that:
 * $DC = CK$

and:
 * $AC = CH$

Therefore from :
 * $KB = 2 \cdot BH$

But we also have:
 * $LH + HB = 2 \cdot BH$

Therefore:
 * $KB = LH + HB$

But:
 * $MNO = CG$

Therefore:
 * $HF = BG$

and as $CD = DG$:
 * $BG = CD \cdot DB$

Also:
 * $HF = CB^2$

Therefore:
 * $CD \cdot DB = CB^2$

Therefore:
 * $DC : CB = CB : BD$

But:
 * $DC > CB$

therefore:
 * $CB > BD$

Therefore, when $CD$ is cut in extreme and mean ratio, the greater segment is $CB$.