We have, $\theta_1+\theta_2=\frac{\pi}{2}$ and $\sqrt{3}(B E)=4 A B$ Let $A B=x$ unit $\begin{aligned} & A C=x \tan \theta_1 \\\\ & E D=x \tan \theta_2 \\\\ & B E=B D+D E \end{aligned}$ $\begin{array}{rlrl} & \Rightarrow \frac{4}{\sqrt{3}} x =x\left(\tan \theta_1+\tan \theta_2\right) {[\because \sqrt{3} B E=4 A B]} \\\\ & \Rightarrow \frac{4}{\sqrt{3}}=\tan \theta_1+\tan \left(\frac{\pi}{2}-\theta_1\right) {\left[\because \theta_1+\theta_2=\frac{\pi}{2}\right]} \end{array}$ $\begin{aligned} & \Rightarrow \tan \theta_1+\cot \theta_1=\frac{4}{\sqrt{3}}=\sqrt{3}+\frac{1}{\sqrt{3}} \\\\ & \Rightarrow \tan \theta_1=\sqrt{3} \text { or } \theta_1=\frac{\pi}{3} \text { and } \theta_2=\frac{\pi}{6} \\\\ & \operatorname{or} \theta_1=\frac{\pi}{6} \text { and } \theta_2=\frac{\pi}{3} \end{aligned}$ $\because \frac{\theta_2}{\theta_1}$ is largest $\therefore \theta_1=\frac{\pi}{6} \text { and } \theta_2=\frac{\pi}{3}$ $\begin{aligned} & \text { Area of } \triangle C A B=\frac{1}{2} \times x \times x \tan \theta_1 \\\\ & \Rightarrow \frac{x^2 \tan \theta_1}{2}=2 \sqrt{3}-3 \\\\ & \Rightarrow x^2=\frac{2(2 \sqrt{3}-3)}{\tan \frac{\pi}{6}}=12-6 \sqrt{3} \\\\\ & \Rightarrow x=3-\sqrt{3} \end{aligned}$ $\text { Also, } C E=\sqrt{x^2+x^2 \tan ^2 \frac{\pi}{3}}=(3-\sqrt{3}) \times 2=6-2 \sqrt{3}$ Perimeter of $\triangle C E D$ $\begin{aligned} & =C D+D E+C E \\\\ & =(3-\sqrt{3})+(3-\sqrt{3}) \tan \frac{\pi}{3}+6-2 \sqrt{3} \\\\ & =3-\sqrt{3}+3 \sqrt{3}-3+6-2 \sqrt{3}=6 \end{aligned}$