Confusing angle-chasing question 
AB = BC = CD = DE = EF = FG = GA
Find angle GAB. 
Please, I want the correct answer. I know how to solve it, but I am getting confused by the number of triangles in it. I am getting different answer every time I solve it. Please tell the correct answer. Options:   
a. 180/5
b. 60
c. 20
d. 180/7 
 A: let $\angle gab=x$, then we have:1. $\angle acb=x$ (since ab=bc)2. $\angle gfa=x$ (since gf=ga)3. $\angle ebd=2x$ (external angle of $\triangle$abc)4. $\angle edb=2x$ (since bc=cd)5. $\angle fge=2x$ (external angle of $\triangle$agf)6. $\angle feg=2x$ (since ef=fg)7. $\angle efd=3x$ (external angle of $\triangle$aef)8. $\angle edf=3x$ (since ef=de)9. $\angle edc=\angle edf-\angle cdb=x$10. $\angle ecd=3x$ (external angle of $\triangle$acd)11. $\angle ced=3x$ (since de=ed)12. $\angle fed=\angle ced-\angle gef=x$
Finally, looking at sum of angles in $\triangle$def, we get:$x+3x+3x=180$
Solve this to find $x$.
A: Denote the unknown angle at $a$ by $\alpha$. Working along the figure it is easy to express any occurring angle in terms of $\alpha$, using only that the angles at the base of an isosceles triangle are equal and that the sum of the three angles of any triangle is $\pi$. You will find angles $\alpha$,  $2\alpha$, $3\alpha$, $\pi-4\alpha$, and the like. Looking at the triangle $ade$ you will finally obtain an equation that can be solved for $\alpha$.
