$解：(1) 連 接 O C .$
$ \because \widehat{A C}=\widehat{B C}， $
$\therefore \angle A O C=\angle B O C . $
$\because C D \perp O A， C E \perp O B， $
$\therefore C D=C E $
$(2) \because \angle A O B=120^{\circ}， \angle A O C=\angle B O C，$
$ \therefore \angle A O C=\angle B O C=60^{\circ} .$
$ \because \angle C D O=90^{\circ}，$
$ \therefore \angle O C D=30^{\circ} .$
$ \because O C=O A=2， $
$\therefore O D=\frac {1}{2}\ \mathrm {O} C=1 .$
$ \therefore C D=\sqrt{O C^2-O D^2}=\sqrt{3} . $
$\therefore S_{\triangle C D O}=\frac {1}{2}\ \mathrm {O} D \cdot C D=\frac {\sqrt{3}}{2}. $
$同理可得， S_{\triangle C E O}=\frac {\sqrt{3}}{2} . $
$\therefore S_{四邊形DOEC }=S_{\triangle C D O}+S_{\triangle C E O}=\sqrt{3} $
