You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
**[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$CD=4+1=5$
|
