$解：(1)证明：连接 O C$
$\because D C 切 O 于点 C$
$\therefore \angle O C D=90^{\circ}$
$又 \because \angle A C D=120^{\circ}$
$\therefore \angle A C O=\angle A C D-\angle O C D=120^{\circ}-90^{\circ}=30^{\circ}$
$\because O C=O A$
$\therefore \angle A=\angle A C O=30^{\circ}$
$\therefore \angle C O D=60^{\circ}$
$\therefore \angle D=30^{\circ}$
$\therefore C A=D C\ $
$(2) \because \sin \angle D=\frac {O C}{O D}=\frac {O C}{O B+B D}=\frac {O B}{O B+B D}$
$\sin \angle D=\sin 30^{\circ}=\frac {1}{2}$
$\therefore \frac {O B}{O B+10}=\frac {1}{2}\ $
$解得OB =10\ $
$即圆 O 的半径为 10\ $
