Zig Zag Zig Zag

Let $\angle AEB = \angle EBD = ... = \theta$ . We note that: $AE \cos{\theta} = EB$ and $EB \sin{\theta} = ED$ .

$\begin{aligned} \Rightarrow EB \sin{\theta} & = ED \\ AE \cos{\theta}\sin{\theta} & = ED \\ 16 \cos{\theta}\sin{\theta} & = 8 \\ 2 \cos{\theta}\sin{\theta} & = 1 \\ \sin{2\theta} & = 1 \\ \Rightarrow \theta & = 45^\circ \end{aligned}$

Therefore, the similar triangles are isosceles triangles with the horizontal side same length with vertical side. And the vertical height is half of that of the previous one. Therefore, we have:

$x=16+8+4+... = 16(1+\frac {1}{2}+\frac {1}{4}+... )$

$\displaystyle \quad = 16\sum _{n=0} ^\infty {\left( \frac {1} {2} \right)^n } = \dfrac {16}{1-\frac{1}{2}} = 32$

By looking the drawing since EB is perpendicular to AC and ED is equal to 1/2 AE, angle BED equal to 45 degrees so DB = ED =1/2EA.

By doing this procedure for every next segment and adding all of them we get a length equal to the first segment AE=16. Since the solution requires we add-up also AE we get 2 AE = 32.