Commentary on Proposition XII.1

Back to Proposition XII.1.

[Folio 0374 | p. 370]

As, from this point onward, the text of each proposition usually occupies considerable space, I shall generally give in the notes a summary of the argument, to enable it to be followed more easily.

Here we have to prove that a pair of corresponding sides are in the ratio of the corresponding diameters.

Since $\angle\text{s}$ BAE, GFL are equal, and the sides about those angles proportional,

$\triangle\text{s}$ ABE, FGL are equiangular,

so that

$\angle$ AEB $= \angle$ FLG

Hence their equals in the same segments, $\angle\text{s}$ AMB, FNG, are equal. And the right angles BAM, GFN are equal. Therefore $\triangle\text{s}$ ABM, FGN are equiangular, so that

BM $:$ GN $=$ BA $:$ GF.

The duplicates of these ratios are therefore equal, whence

$(\text{polygon }$ ABCDE $) : (\text{polygon }$ FGHKL $)$ $= \text{duplicate ratio of }$ BA $\text{ to }$ GF $= \text{duplicate ratio of }$ BM $\text{ to }$ GN $=$ BM² $:$ GN².