Find Length

Draw line AG // BC; CG ⊥ BC; extend AD so it intersects CD at F. Since ∠DAC = 2β, so FC = 2CG = 2AE = 12.
Since θ = 180° - 45° - a = β + 45°; θ’ = 2β + 45° - β = β + 45°, so θ = θ', ΔED’C = ΔFD’C, & EC = FC = 12

Let $\angle ACB = \theta$ . Drop a altitude from $D$ to $BC$ and draw two straight lines parallel to $BC$ through $A$ and $D$ . We note that $\angle DAG = \angle EDF = \theta$ . Since $\triangle CDF$ is isosceles, $DF = CF$ ; let their length be $\color{#3D99F6} h$ . Then we have:

$\begin{cases} DF = DG + GF & \implies h = a \tan \theta + 6 & ...(1) \\ CF = BC - BF & \implies h = \dfrac 6{\tan \theta} - a & ...(2) \end{cases} \quad \small \color{#3D99F6} \text{where }AG = BF = a$

From $(1) + \tan \theta \times (2): \quad h + h \tan \theta = 12 \implies h = \dfrac {12}{1+\tan \theta}$

Now $CE = CF + EF = h + h \tan \theta = \dfrac {12}{1+\tan \theta} \times (1+\tan \theta) = \boxed {12}$ .

Place the diagram on a coordinate system so that $B$ is at the origin, $C$ is on the $x$ -axis, and $A$ is on the $y$ -axis. Let $\theta = \angle ACB$ .

By trigonometry on $\triangle ABC$ , $BC = \frac{6}{\tan \theta}$ .

Line $CD$ has a slope of $-1$ and an $x$ -intercept (and a $y$ -intercept) of $\frac{6}{\tan \theta}$ , so its equation is $y = -x + \frac{6}{\tan \theta}$ .

Line $AD$ has a slope of $\tan((90° - \theta) + 2\theta - 90°) = \tan(\theta)$ and a $y$ -intercept of $6$ , so its equation is $y = (\tan \theta)x + 6$ .

Lines $AD$ and $CD$ intersect at $D(t, u)$ , where $t = \frac{6(1 - \tan \theta)}{\tan \theta (1 + \tan \theta)}$ and $u = \frac{12}{1 + \tan \theta}$ .

Line $AC$ has a slope of $\tan(180° - \theta) = -\tan \theta$ .

Since $DE \perp AC$ , $DE$ has a slope of $\frac{1}{\tan \theta}$ , and since it goes through $(t, u)$ , its equation is $y = \frac{1}{\tan \theta}(x - t) + u$ .

Solving $0 = \frac{1}{\tan \theta}(x - t) + u$ gives $x = \frac{6}{\tan \theta} - 12 = BE$ .

Therefore, $CE = BC - BE = \frac{6}{\tan \theta} - (\frac{6}{\tan \theta} - 12) = \boxed{12}$ .

This solution is inspired by @Chew-Seong Cheong. Drop an altitude from $D$ to $BC$ and draw a straight line parallel to $BC$ through $A.$ $\begin{aligned} &\left\{ \begin{aligned} \triangle{ADG}\sim\triangle{CAB}\implies\frac{DG}{AB}=\frac{AG}{BC}\\ \triangle{DEF}\sim\triangle{CAB}\implies\frac{EF}{AB}=\frac{DF}{BC} \end{aligned} \right.\\\\ \implies & \frac{DG+EF}{AB}=\frac{AG+DF}{BC}=\frac{BF+CF}{BC}=1\\\\ \implies & CE=CF+EF=DF+EF=FG+DG+EF=AB+AB=\boxed{12}. \end{aligned}$

Let $\overline{DH}=y$ . As $\overline{AH}$ bisects $\angle DAG$ and $\overline{AH} \perp \overline{DG}$ , $\overline{DH} \cong \overline{HG}$ . As $\overline{HF}=6$ , then $\overline{FG}=6-y$ . As $\bigtriangleup DFC$ is isosceles, $\overline{FC}=6+x$ .

Then, as $\bigtriangleup EFD \sim \bigtriangleup GFC$ we have:

$\frac{x-6-y}{6+y}=\frac{6-y}{6+y}$ $x=12$