Поскольку АВ=ВС, то угол ABC равен углу ACB, так что произведем замену углов:

АВС=2α, ВСА=2β. Тогда ВCD=5α, СDE=5β.

Возьмём AC=2R. Тогда BC=2Rsinα, AD=2Rsinβ, BD=2Rsin$\alpha +\beta$.

Воспользуемся теоремой синусов для треугольника BCD:

R/sin$5\alpha$=2Rsin$\alpha +\beta$/sin$5\alpha$.

Упростим:

1/sin$5\alpha$=2sin$\alpha +\beta$

1/sin$5\alpha$=2sinαcosβ+2cosαsinβ

$sin\alpha cos\beta +cos\alpha sin\beta$^2=1/4$1-co{s}^2(5\alpha )$

sin^2$\alpha +\beta$=1/4$1-si{n}^2(5\alpha )$

sin^2$\alpha +\beta$=1-cos^2$5\alpha$

cos^2$\alpha +\beta$+sin^2$\alpha +\beta$=1-cos^2$5\alpha$

Так как BD и CE делятся пополам:

sin$\alpha +\beta$/sin5β=3/2sin5α

sin$\alpha +\beta$=3/2cosβ

Тогда sinβ/sin$\alpha +\beta$=5/3

cosβ=5sin$\alpha +\beta$/3

$1-co{s}^2(\alpha +\beta )$$1-si{n}^2(5\alpha )$=5^2/3^2

cos^2$\alpha +\beta$+cos^2$5\alpha$=1-25/9

5cos^2$5\alpha$=4

R=2/5sin$5\alpha$

Подставим это в AC=2R:

AC=4sin$5\alpha$/5

$СА\cdot СB$$СD\cdot DE$=1

4sin$5\alpha$/5·2R·2Rsin$\alpha +\beta$$2R\cdot 2Rsin(5\alpha )$=1

4sin$5\alpha$/5·2/5sin$5\alpha$·2/5sin5α$4/5sin2\alpha$

sinα\sinβ=1/5

sinα=sqrt$5$/5

sinβ=sqrt$5$/5

sin$\alpha +\beta$=3sqrt$2$/10

cos$\alpha +\beta$=4/5

cosβ=5sqrt$2$/6

AD=2Rsinβ=4sqrt$5$/3

CD=2Rsin5β=2√5

We also need to find the length of other segments. Let's find angle AJH by finding the angle BHJ first. Since AB = BC, ∠AHB = ∠BJD = 180 - ∠ABC. Hence, ∠FBJ = ∠BJD - ∠FBD = 180 - ∠ABC - ∠FBD.

Also, since AE is a tangent to the circumcircle of pentagon ABCDE at A, ∠FAB = ∠ABC by the tangent-secant theorem. Thus, ∠FBD = 180 - ∠ABC.

Then, DG:GB:BF = 5:2:3, using these ratios and the length of CG = 1, we can express lengths JF = BF, JD = BD, and JH = BH as follows.
CG:GB = 1:2. Since GD:GB = 5:2, JD = $5/2$ * GB = 5, DG = 5 - 1 = 4; BC = AB, FC = BF.
Use the Law of Sines to find AF. As we found the measures of sin$\alpha +\beta$ and cos$\alpha +\beta$ as well from the above calculations.
The length of segment AF:
AF = FC / sin$\angle CFA$ = 20/3.