$解：(1) 如圖， 連接 A D .$
$\because A B 是 \odot O 的直徑，\therefore \angle A D B=90^{\circ}.$
$\therefore A D \perp B C.$
$又 \because D C=B D，$
$\therefore A B=A C \quad$
$(2) 如圖， 連接 O D .\because D E \perp A C，$
$\therefore \angle C E D=90^{\circ} .\because O A=O B， D C=B D，$
$\therefore O D 是 \triangle A B C 的中位線.$
$\therefore O D / / A C.\therefore \angle O D E=\angle C E D=90^{\circ}\therefore D E \perp O D.$
$又 \because O D 是 \odot O 的半徑，$
$\therefore D E 是 \odot O 的 切線$
$(3) \because A B=A C， \angle B A C=60^{\circ}，$
$\therefore \triangle A B C 是等邊三角形.\therefore B C=A B.$
$\because \odot O 的半徑為6\therefore A B=B C=12\ $
$.\therefore C D=\frac {1}{2}\ \mathrm {B} C=6 .$
$\because \angle C=60^{\circ}， D E \perp A C，$
$\therefore \angle C D E=30^{\circ} .$
$\therefore C E=\frac {1}{2}\ \mathrm {C} D=3 .$
$\therefore D E=\sqrt{C D^2-C E^2}=3 \sqrt{3}$
