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#### 口诀
< font color = 'red' size = 4 >< b > 等线段，共端点，辅助线，用旋转</ b ></ font >
** 解惑**：
** $Q:$为什么要用旋转？**
$A:$旋转后，由于等线段，可以完美边重合，可以构建全等三角形，使得问题转化为同一个三角形中的条件，问题就简化了。
如图：
延长$CB$至$E$，使得$BE=CD$
$AC=AE,BE=CD,\angle ABE+\angle ABC=180^{\circ}$
$\because \angle ABC+\angle ADC=180^{\circ}$
$\therefore \angle ABE=\angle ADC$
$\therefore \triangle ABE \cong ADC$
$\therefore \angle 1=\angle 2,AC=AE$
双 $\because \angle 1 +\angle 3=\angle 2+ \angle 3=90^{\circ}$
$\therefore \triangle ACE$是等腰直角三角形
而$S_{\triangle ACE}=$四边形面积
$=\frac{1}{2}AC\cdot AE=18$
$=\frac{1}{2} AC ^2=18$
$\therefore AC=6$
