1. 如图,在$\triangle ABC$中,以$AB为直径的\odot O分别交AC于点D$,交$BC于点E$,连接$ED$,且$ED= EC$.
(1)求证：$AB= AC$；
(2)若$\frac{AD}{DC}= \frac{3}{2}$,求$\frac{AB}{ED}$的值.

解: (1) $\because ED = EC$,
$\therefore \angle EDC = \angle C$.
$\because \angle EDC + \angle ADE = 180^{\circ}$,
$\angle ABE + \angle ADE = 180^{\circ}$,
$\therefore \angle EDC = \angle ABE$,
$\therefore \angle ABE = \angle C$, $\therefore AB = AC$;
(2) 连接 $AE$, $BD$. $\because \frac{AD}{DC} = \frac{3}{2}$,
$\therefore$ 设 $AD = 3a$,
则 $DC = 2a$, $AB = AC = 5a$,
$\because AB$ 为 $\odot O$ 的直径,
$\therefore \angle AEB = \angle ADB = \angle BDC = 90^{\circ}$.
在 $Rt\triangle ABD$ 中,
$BD = \sqrt{AB^{2} - AD^{2}} = 4a$.
在 $Rt\triangle CBD$ 中,
$BC = \sqrt{BD^{2} + CD^{2}} = 2\sqrt{5}a$.
又 $\because AB = AC$,
$\therefore BE = CE = \frac{1}{2}BC = \sqrt{5}a$,
$\therefore ED = EC = \sqrt{5}a$, $\therefore \frac{AB}{ED} = \sqrt{5}$.

2. 如图,在$\triangle ABC$中,$AB= AC$,以$AB为直径的\odot O交BC于点D$,交$CA的延长线于点E$.
(1)求证：$BD= CD$；
(2)连接$ED$,若$AC= 5$,$ED= 4$,求$AE$的长.

解: (1) 连接 $AD$.
$\because AB$ 为 $\odot O$ 的直径,
$\therefore \angle ADB = 90^{\circ}$.
$\because AB = AC$, $\therefore BD = DC$;
(2) 连接 $BE$, 则 $\angle AEB = 90^{\circ}$,
$\because BD = CD$, $\therefore BC = 2ED = 8$.
设 $AE = x$, 则 $CE = x + 5$.
在 $Rt\triangle ABE$ 和 $Rt\triangle CBE$ 中,
$\because AB^{2} - AE^{2} = BE^{2} = BC^{2} - CE^{2}$,
$\therefore 5^{2} - x^{2} = 8^{2} - (x + 5)^{2}$, $\therefore x = \frac{7}{5}$,
即 $AE$ 的长为 $\frac{7}{5}$.

3. 如图,在$\triangle ABC$中,$BA= BC$,以$AB为直径的\odot O分别交AC$,$BC于点D$,$E$,$BE= 4CE$,$AD= \sqrt{10}$.
(1)求证：$AD= CD$；
(2)求$S_{\triangle ABC}$.

解: (1) 连接 $BD$. $\because AB$ 为 $\odot O$ 的直径, $\therefore \angle ADB = 90^{\circ}$.
$\because BA = BC$, $\therefore AD = CD$;
(2) 连接 $AE$, 则 $\angle AEB = 90^{\circ}$.
设 $CE = x$, $BE = 4x$,
则 $BC = 5x = AB$,
在 $Rt\triangle AEB$ 中,
$AE = \sqrt{AB^{2} - BE^{2}} = 3x$,
在 $Rt\triangle ACE$ 中,
$(3x)^{2} + x^{2} = (2\sqrt{10})^{2}$,
$\therefore x = 2$, $\therefore BC = 10$, $AE = 6$,
$\therefore S_{\triangle ABC} = \frac{1}{2}BC \cdot AE = 30$.