$证明:(1)∵ {\widehat{AD}}={\widehat{BD}},∴ ∠ACD=∠ECB$
$∵ {\widehat{AC}}={\widehat{AC}},∴∠ADC=∠EBC.∴△ACD∽△ECB$
$(2)过点B作 B ⊥CD于点\ $H
$∵AB是⊙O的直径,∴∠ACB=∠ADB= 90°$
$在Rt△ACB 中, AB=\sqrt{BC^{2}+ AC^{2}}=\sqrt{10}$
$∵{\widehat{AD}}={\widehat{BD}}, ∴∠ACD=∠BCD =45°° ∴ ∠ABD = ∠BAD=45°$
$∴△ABD为等腰直角三角形,∴易得BD= \frac{\sqrt{2}}{2}AB=\frac{\sqrt{2}}{2}×\sqrt{10}=\sqrt{5}$
$在Rt△BCH中，∵∠BCH=45°, ∴易得CH=BH=\frac{\sqrt{2}}{2}BC=\frac{\sqrt{2}}{2}$
$在 Rt△BDH中,DH= \sqrt{BD^{2}-BH^{2}}=\frac{3\sqrt{2}}{2},∴ CD=CH+DH=2\sqrt{2}$
$∵△ACD∽△ECB,∴\frac{CA}{CE}=\frac{CD}{CB},∴CE=\frac{3\sqrt{2}}{4}$
