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Commentary on Proposition XIII.9¶

BC is the side of a regular decagon inscribed in the circle; CD is the side of the inscribed regular hexagon, and is therefore equal to the radius BE or EC.

Therefore, in order to prove our theorem, we have only to show that the triangles EBC, DBE are similar.

[Folio 0461 | p. 457]

Since BC is the side of a regular decagon,

$(\text{arc }$ BCA $) = 5\,(\text{arc }$ BC $)$, so that $(\text{arc }$ CFA $) = 4\,(\text{arc }$ BC $)$, whence $\angle$ CEA $= 4\,\angle$ BEC.

But $\angle$ CEA $= 2\,\angle$ ECB.

Therefore $\angle$ ECB $= 2\,\angle$ BEC $\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\text{(1)}$.

But, since CD $= $ CE, $\angle$ CDE $= \angle$ CED, so that $\angle$ ECB $= 2\,\angle$ CDE.

It follows from (1) that $\angle$ BEC $= \angle$ CDE.

Now, in the $\Delta\text{s}$ EBC, DBE,

$\angle$ BEC $= \angle$ BDE,

and $\angle$ EBC is common,

so that $\angle$ ECB $= \angle$ DEB, and

$\Delta\text{s}$ EBC, DBE are similar.

Hence

DB $:$ BE $= $ EB $:$ BC,

$(\text{rect. }$ DB, BC $) = (\text{sq. on }$ EB $)$ $= (\text{sq. on }$ CD $)$,

and DB is divided at C in extreme and mean ratio.

To find the side of the decagon algebraically in terms of the radius we have, if $x$ be the side required,

$$(r + x)x = r^2,$$

whence

$$x = \frac{r}{2}(\sqrt{5} - 1).$$