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** [前导知识：一线三等角模型 ](https://baijiahao.baidu.com/s?id=1690759343287283314&wfr=spider&for=pc )**
** 分析题意**：
$\because \angle ADC=45^{\circ}+\angle 1$
$\because \angle ADC=45^{\circ}+\angle 2$
$\therefore \angle 1=\angle 2$
此时，构造相似三角形，通过比例关系解题就是 ** 关键**
因为$\angle 1$在$\triangle EDC$中，同时知道$EC$长度为$\sqrt{5}$,我们需要构造出一个和 $\triangle ABD$相似的三角形，所以过$E$引$EF$交$CD$于$F$,使得$\angle EFD=45^{\circ}$
则$\triangle EDF \sim \triangle ABD$
$\therefore \frac{AD}{AB}=\frac{DE}{FD}$
$\because DE=\sqrt{2} AD$
$\therefore FD=\sqrt{2}AB=\sqrt{2}*2\sqrt{2}=4$
算出$DF$后，下面需要继续求解$FC$,才能算出$CD$的长度。
继续观察发现，
$$
\left\{\begin{matrix}
\angle C=\angle C & \\
\angle CFE=135^{\circ} & \\
\angle CDE=135^{\circ} &
\end{matrix}\right.
$$
$\therefore \triangle CEF \sim \triangle CDE$
设$CF=x$
$\therefore \frac{x}{CE}=\frac{CE}{CD}$
$x^2+4x=5$
解方程：$x_1=1,x_2=-5$,因为$x>0$,$x_2=-5$舍掉，最终$x=1$,所以$CD=4+1=5$
