解:$(1)$设$AC$长为$x$
$ x^2=(1-x)×1$
$ $解得$x_{1}=\frac {\sqrt {5}-1}2,$$x_{2}=\frac {-\sqrt {5}-1}2($不合题意,舍去)
∴$ AC$的长为$\frac {\sqrt {5}-1}2$
$ (2) $由$(1)$得$AC=\frac {\sqrt {5}-1}2AB$
∴$ AD=\frac {\sqrt {5}-1}2AC=\frac {3-\sqrt {5}}2$
∴$ AD$的长度为$\frac {3-\sqrt {5}}2$
$ (3)AE=\frac {\sqrt {5}-1}2AD=\sqrt {5}-2$
∴$ AE$的长度为$\sqrt {5}-2$
规律:$ $所求长度恰是条件等式右边较长线段长度的$\frac {\sqrt {5}-1}2$倍