Commentary on Proposition X.32¶
Back to Proposition X.32.
Euclid takes three straight lines of the form $\rho$, $\rho\sqrt{\lambda}$, $\rho\sqrt{1-k^2}$, takes the mean proportional $\rho\lambda^{\frac{1}{4}}$ between the first two ………………..(1), and then finds $x$ such that
$$\rho\lambda^{\frac{1}{4}} : \rho\sqrt{\lambda} = \rho\sqrt{1-k^2} : x \text{ .........................................(2),}$$
whence $x = \rho\lambda^{\frac{1}{4}}\sqrt{1-k^2}$, and the straight lines $\rho\lambda^{\frac{1}{4}}$, $\rho\lambda^{\frac{1}{4}}\sqrt{1-k^2}$ satisfy the given conditions.
Now ($\alpha$) $\rho\lambda^{\frac{1}{4}}$ is medial.
($\beta$) We have, from (1) and (2),
$$\rho : \rho\sqrt{1-k^2} = \rho\lambda^{\frac{1}{4}} : x \text{ .........................................(3),}$$
whence $x \mathbin{\sim} \rho\lambda^{\frac{1}{4}}$ ; and $x$ is therefore medial and $\mathbin{\sim} \rho\lambda^{\frac{1}{4}}$.
($\gamma$) $x\ .\ \rho\lambda^{\frac{1}{4}} = \rho\sqrt{\lambda}\ .\ \rho\sqrt{1-k^2}$. But the latter is medial ; [x. 21] therefore $x\ .\ \rho\lambda^{\frac{1}{4}}$, or $\rho\lambda^{\frac{1}{4}}\ .\ \rho\lambda^{\frac{1}{4}}\sqrt{1-k^2}$, is medial.
Lastly ($\delta$) $\rho$, $\rho\sqrt{1-k^2}$ have the remaining property in the enunciation ; therefore $\rho\lambda^{\frac{1}{4}}$, $\rho\lambda^{\frac{1}{4}}\sqrt{1-k^2}$ have it also. [x. 14]
(Euclid has not the assistance of symbols to prove the proportion (3) above. He therefore uses the lemmas $ab : bc = a : c$ and $d^2 : de = d : e$ to deduce from the relations
$$\left.\begin{array}{r} ab = d^2 \\ d : b = c : e \end{array}\right\}$$
and that $a : c = d : e$.)
The straight lines $\rho\lambda^{\frac{1}{4}}$, $\rho\lambda^{\frac{1}{4}}\sqrt{1-k^2}$ may take any of the following forms according as the straight lines first taken are (1) $a$, $\sqrt{B}$, $\sqrt{a^2-c^2}$, (2) $\sqrt{A}$, $\sqrt{B}$, $\sqrt{A-k^2A}$, (3) $\sqrt{A}$, $b$, $\sqrt{A-k^2A}$.
$$\begin{array}{lcc} (1) & \sqrt{a\sqrt{B}}, & \dfrac{\sqrt{B(a^2-c^2)}}{\sqrt{a\sqrt{B}}} ; \\ (2) & \sqrt[4]{AB}, & \dfrac{\sqrt{B(A-k^2A)}}{\sqrt[4]{AB}} ; \\ (3) & \sqrt{b\sqrt{A}}, & \dfrac{b\sqrt{A-k^2A}}{\sqrt{b\sqrt{A}}}. \end{array}$$
If the other conditions are the same, but the square on the first medial straight line is to exceed the square on the second by the square on a straight line incommensurable with the first, we begin with the three straight lines $\rho$, $\rho\sqrt{\lambda}$, $\dfrac{\rho}{\sqrt{1+k^2}}$, and the medial straight lines are
$$\rho\lambda^{\frac{1}{4}},\quad \dfrac{\rho\lambda^{\frac{1}{4}}}{\sqrt{1+k^2}}.$$
The possible forms are even more various in this case owing to the more various forms that the original lines may take, e.g.
$$\begin{array}{llll} (1) & a, & \sqrt{B}, & \sqrt{a^2-C} ; \\ (2) & \sqrt{A}, & b, & \sqrt{A-c^2} ; \\ (3) & \sqrt{A}, & b, & \sqrt{A-C} ; \\ (4) & \sqrt{A}, & \sqrt{B}, & \sqrt{A-c^2} ; \\ (5) & \sqrt{A}, & \sqrt{B}, & \sqrt{A-C} ; \end{array}$$
the medial straight lines corresponding to these being
$$\begin{array}{lcc} (1) & \sqrt{a\sqrt{B}}, & \dfrac{\sqrt{B(a^2-C)}}{\sqrt{a\sqrt{B}}} ; \\ (2) & \sqrt{b\sqrt{A}}, & \dfrac{b\sqrt{A-c^2}}{\sqrt{b\sqrt{A}}} ; \\ (3) & \sqrt{b\sqrt{A}}, & \dfrac{b\sqrt{A-C}}{\sqrt{b\sqrt{A}}} ; \\ (4) & \sqrt[4]{AB}, & \dfrac{\sqrt{B(A-c^2)}}{\sqrt[4]{AB}} ; \\ (5) & \sqrt[4]{AB}, & \dfrac{\sqrt{B(A-C)}}{\sqrt[4]{AB}}. \end{array}$$
LEMMA.¶
Let ABC be a right-angled triangle having the angle A right, and let the perpendicular AD be drawn ;
I say that the rectangle CB, BD is equal to the square on BA, the rectangle BC, CD equal to the square on CA, the rectangle BD, DC equal to the square on AD, and, further, the rectangle BC, AD equal to the rectangle BA, AC.
And first that the rectangle CB, BD is equal to the square on BA.
I say that the rectangle BC, AD is also equal to the rectangle BA, AC.