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

Proof

 * Euclid-XIII-2.png

Suppose that $BC = 2 \cdot CA$.

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

Therefore:
 * $BC^2 + CA^2 = 5 CA^2$

But by hypothesis:
 * $BA^2 = 5 CA^2$

Therefore by :
 * $BA^2 = BC^2 + CA^2$

which is impossible.

Therefore:
 * $BC \ne 2 \cdot CA$

Similarly it can be shown that it is not the case that:
 * $BC < 2 \cdot CA$

as it leads to a much greater absurdity.

Therefore:
 * $BC > 2 \cdot CA$