In the regular pentagon $$ABCDE$$, the perpendicular at $$C$$ to $$CD$$ meets $$AB$$ at $$F$$. Prove that $$AE + AF = BE$$.

Take the side of the pentagon as $$AB = BC = AE = 1$$; each interior angle is $$108^\circ$$. Since $$\angle BCD = 108^\circ$$ and $$\angle FCD = 90^\circ$$, we get $$\angle BCF = 18^\circ$$, and in $$\triangle BCF$$, $$\angle FBC = 108^\circ$$, so $$\angle BFC = 54^\circ$$. By the sine rule $$BF = \frac{\sin 18^\circ}{\sin 54^\circ} = \frac{(\sqrt{5} - 1)/4}{(\sqrt{5} + 1)/4} = \frac{3 - \sqrt{5}}{2}$$, so $$AF = 1 - BF = \frac{\sqrt{5} - 1}{2}$$. In isosceles $$\triangle ABE$$, $$\angle ABE = \angle AEB = 36^\circ$$, so the sine rule gives $$BE = \frac{\sin 108^\circ}{\sin 36^\circ} = \frac{\sin 72^\circ}{\sin 36^\circ} = 2\cos 36^\circ = \frac{\sqrt{5} + 1}{2}$$. Hence $$AE + AF = 1 + \frac{\sqrt{5} - 1}{2} = \frac{\sqrt{5} + 1}{2} = BE$$.