Kind of a harder one. Full picture

\

Aha! CBD ~ CGA!

Both are 30-60-90.

1. Set CG = $x$, and set CB = $y$

2. CD is $x+1$

3. Based on 30-60-90 properties, $\frac{x+1}{2}=y$

4. Write a similarity proportion

$\frac{y}{x+1}=\frac{x}{y+13}$

So now we have two equations...

Have fun solving!

Here is my solution :

https://imgur.com/a/hhC4nVl

I just used some basic geometry theories like Equilateral Triangle, Pythago, etc.

Hope it can help :)

Triangles  BCD, ACG, and GDN are similar;  all of them are  30-60-90

Let's denote the intersection point of AF and BD with a leter N

BN = 13 * tan(30°) = 7.505

DN = 1 / cos (30°)  = 1.155

BD = BN + DN = 8.66

CD = BD / cos (30°) = 10

Circle radius r = CD / 2 = 5

EG = sqrt [ r² - ( r - 1 )² ] = 3

CG = CD - GD = 9

AG = CG / tan (30°)  = 15.588

EA = AG - EG = 12.588    

Thanks guys,

That is really impressive.

I followed your logic and then drew the pic.

Here it is.

I have coloured the 3 similar triangles that are used in the solution.