CHAPTER XIV¶
On Its Fifth Marvelous Effect
If a quantity be divided according to our said proportion, the sum of the square of the lesser part and the square of the entire undivided quantity will always be triple the square of the greater part ([XIII.4] [Campanus XIII.5]):
$$b^2 + (a + b)^2 = 3a^2$$
Numerical Demonstration
For example, let $10$ be the divided quantity, where, as we have said, one part is $15 - \sqrt{125}$ (namely the lesser, $b$) and the other is $\sqrt{125} - 5$ (namely the greater, $a$).
I say that when the square of $15 - \sqrt{125}$ is added to the square of $10$ (the whole quantity), their sum will be triple—that is, three times as great as—the square of the greater part, namely of $\sqrt{125} - 5$:
The square of the lesser part ($15 - \sqrt{125}$) is:
$$b^2 = (15 - \sqrt{125})^2 = 225 + 125 - 30\sqrt{125} = 350 - \sqrt{112500}$$
The square of the whole quantity ($10$) is $100$, which, added to $350 - \sqrt{112500}$, makes:
$$b^2 + (a + b)^2 = 450 - \sqrt{112500}$$
for the said sum.
Meanwhile, the square of the greater part ($\sqrt{125} - 5$) is:
$$a^2 = (\sqrt{125} - 5)^2 = 150 - \sqrt{12500}$$
which is manifestly one-third ($\frac{1}{3}$) of the said sum! For when $150 - \sqrt{12500}$ is multiplied by $3$, it makes:
$$3a^2 = 3 \times 150 - \sqrt{9 \times 12500} = 450 - \sqrt{112500}$$