21.（本小题满分12分）
如图，△ABC内接于⊙O，AB=AC，D是$\overset{\frown}{AC}$上一点，过点C作CE//AD，交BD于点E.
(1)求证：$CD=DE$.
(2)若$AB=10$，$BC=4\sqrt{5}$，$BE=6$.
①求AD的长.
②求CD的长.

21. (1) 证明：$\because AD// CE,$
$\therefore ∠ADB=∠DEC.$
$\because ∠ADB=∠ACB,$
$\therefore ∠ACB=∠DEC.$
$\because ∠BAC=∠BDC,\ ∠BAC+∠ABC+∠ACB=180°,\ ∠BDC+∠DEC+∠DCE=180°,$
$\therefore ∠ABC=∠DCE.$
$\because AB=AC,$
$\therefore ∠ABC=∠ACB.$
$\therefore ∠DCE=∠ABC=∠ACB=∠DEC.$
$\therefore CD=DE.$
(2) 解：① $\because ∠ACB=∠DCE,$
$\therefore ∠ACB-∠ACE=∠DCE-∠ACE.$
$\therefore ∠BCE=∠ACD.$
又 $\because ∠DBC=∠DAC,$
$\therefore △ BCE ∽ △ ACD.$
$\therefore \dfrac{BC}{AC}=\dfrac{BE}{AD}.$
$\because AB=AC=10,\ BC=4\sqrt{5},\ BE=6,$
$\therefore \dfrac{4\sqrt{5}}{10}=\dfrac{6}{AD}.$
$\therefore AD=3\sqrt{5}.$
② 如图，过点 $A$ 作 $AF⊥ BC$，垂足为 $F$，作 $AG⊥ BD$，垂足为 $G.$

$\because AB=AC,\ AF⊥ BC,$
$\therefore CF=\dfrac{1}{2}BC=2\sqrt{5}.$
在 $\mathrm{Rt}△ACF$ 中，$\cos ∠ACF=\dfrac{CF}{AC},$
在 $\mathrm{Rt}△ADG$ 中，$\cos ∠ADG=\dfrac{DG}{AD}.$
$\because ∠ACF=∠ADG,$
$\therefore \cos ∠ACF=\cos ∠ADG.$
$\therefore \dfrac{CF}{AC}=\dfrac{DG}{AD}.$
$\therefore \dfrac{2\sqrt{5}}{10}=\dfrac{DG}{3\sqrt{5}}.$
$\therefore DG=3.$
$\therefore AG=\sqrt{AD^2-DG^2}=\sqrt{(3\sqrt{5})^2-3^2}=6.$
在 $\mathrm{Rt}△ABG$ 中，$AB=10,$
$\therefore BG=\sqrt{AB^2-AG^2}=\sqrt{10^2-6^2}=8.$
$\therefore BD=BG+DG=8+3=11.$
$\therefore CD=DE=BD-BE=11-6=5.$