解:?$(1)$?設(shè)?$AC$?長(zhǎng)為?$x$?
?$ x^2=(1-x)×1$?
?$ $?解得?$x_{1}=\frac {\sqrt {5}-1}2,$??$x_{2}=\frac {-\sqrt {5}-1}2($?不合題意,舍去)
∴?$ AC$?的長(zhǎng)為?$\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$?的長(zhǎng)度為?$\frac {3-\sqrt {5}}2$?
?$ (3)AE=\frac {\sqrt {5}-1}2AD=\sqrt {5}-2$?
∴?$ AE$?的長(zhǎng)度為?$\sqrt {5}-2$?
規(guī)律:?$ $?所求長(zhǎng)度恰是條件等式右邊較長(zhǎng)線(xiàn)段長(zhǎng)度的?$\frac {\sqrt {5}-1}2$?倍