Commentary on Proposition XIII.10

Back to Proposition XIII.10.

ABCDE being a regular pentagon inscribed in a circle, and AG the diameter through A, it follows that $(\text{arc }$ CG $) = (\text{arc }$ GD $), and CG, GD are sides of an inscribed regular decagon.

FHK being drawn perpendicular to AB, it follows, by I. 26, that $\angle\text{s}$ AFK, BFK are equal, and BK, KA are sides of the regular decagon.

Similarly it may be proved that, FLM being perpendicular to AK, AL $= $ LK, and $(\text{arc }$ AM $) = (\text{arc }$ MK $).

The main facts to prove are that (1) the triangles ABF, FBN are similar, and (2) the triangles ABK, AKN are similar.

  1. $2\,(\text{arc }$ CG $) = (\text{arc }$ CD $)$ $= (\text{arc }$ AB $)$ $= 2\,(\text{arc }$ BK $), or $(\text{arc }$ CG $) = (\text{arc }$ BK $) = (\text{arc }$ AK $)$ $= 2\,(\text{arc }$ KM $). And $(\text{arc }$ CB $) = 2\,(\text{arc }$ BK $). Therefore, by addition, $(\text{arc }$ BCG $) = 2\,(\text{arc }$ BKM $). Therefore $\angle$ BFG $= 2\,\angle$ BFN. But $\angle$ BFG $= 2\,\angle$ FAB, so that $\angle$ FAB $= \angle$ BFN. Hence, in the $\Delta\text{s}$ ABF, FBN, $\angle$ FAB $= \angle$ BFN, and $\angle$ ABF is common; therefore $\angle$ AFB $= \angle$ BNF, and $\Delta\text{s}$ ABF, FBN are similar.

  2. Since AL $= $ LK, and the angles at L are right, AN $= $ NK, and $\angle$ NKA $= \angle$ NAK $= \angle$ KBA. Hence, in the $\Delta\text{s}$ ABK, AKN, $\angle$ ABK $= \angle$ AKN, and $\angle$ KAN is common; whence the third angles are equal; therefore the triangles ABK, AKN are similar.

Now from the similarity of $\Delta\text{s}$ ABF, FBN it follows that AB $:$ BF $= $ BF $:$ BN, or $(\text{rect. }$ AB, BN $) = (\text{sq. on }$ BF $)$.

And, from the similarity of ABK, AKN, BA $:$ AK $= $ AK $:$ AN, or $(\text{rect. }$ BA, AN $) = (\text{sq. on }$ AK $)$.

Therefore, by addition, $(\text{rect. }$ AB, BN $) + (\text{rect. }$ BA, AN $) = (\text{sq. on }$ BF $) + (\text{sq. on }$ AK $), that is, $(\text{sq. on }$ AB $) = (\text{sq. on }$ BF $) + (\text{sq. on }$ AK $)$.

If $r$ be the radius of the circle, we have seen (XIII. 9, note) that AK $= \frac{r}{2}(\sqrt{5} - 1)$.

Therefore

$$\begin{aligned} (\text{side of pentagon})^2 &= r^2 + \frac{r^2}{4}(6 - 2\sqrt{5}) \\ &= \frac{r^2}{4}(10 - 2\sqrt{5}), \end{aligned}$$

so that

$$(\text{side of pentagon}) = \frac{r}{2}\sqrt{10 - 2\sqrt{5}}.$$