9. 如图,点A,B,C在$\odot O$上,直径$AD=4$,$∠ ABC=∠ DAC$,则$AC=$______.

$2\sqrt{2}$

10. 如图,AB 是$\odot O$的直径,C 是$\odot O$上一点,$∠ ACD=30°$.
(1) 求$∠ DAB$的度数.
(2) 过点 D 作$DE ⊥ AB$,垂足为 E,DE 的延长线交$\odot O$于点 F.若$AB=4$,求 DF 的长.

解：（1）如图，连接$BD。$ $\because \angle ACD = 30^{\circ}，$$\therefore \angle B=\angle ACD = 30^{\circ}。$ $\because AB$是$\odot O$的直径，$\therefore \angle ADB = 90^{\circ}。$ $\therefore \angle DAB = 90^{\circ}-\angle B = 60^{\circ}。$ （2）$\because \angle ADB = 90^{\circ}，$$\angle B = 30^{\circ}，$$AB = 4，$ $\therefore AD=\frac{1}{2}AB = 2。$ $\because \angle DAB = 60^{\circ}，$$DE\perp AB，$且$AB$是$\odot O$的直径， $\therefore \angle ADE = 30^{\circ}，$$DE = EF。$ $\therefore AE=\frac{1}{2}AD = 1。$ $\therefore DE=\sqrt{AD^{2}-AE^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}。$ $\therefore DF = DE + EF = 2DE = 2\sqrt{3}。$ ;

12. 已知$A,B,C,D$是$\odot O$上的四个点.
（1）如图①,若$∠ ADC=∠ BCD=90°,AD=CD$,求证:$AC⊥ BD$;
（2）如图②,若$AC⊥ BD$,垂足为$E$,$AB=2,DC=4$,求$\odot O$的半径.

第12题

解：（1）$\because \angle ADC=\angle BCD = 90^{\circ}，$ $\therefore AC，$$BD$是$\odot O$的直径，且交点为圆心$O。$ $\because AD = CD，$$AO = CO，$ $\therefore AC\perp BD。$ （2）连接$CO$并延长，交$\odot O$于点$K，$连接$DK，$$BC，$则$\angle KDC = 90^{\circ}。$ $\therefore \angle K+\angle KCD = 90^{\circ}。$ $\because AC\perp BD，$$\therefore \angle ACB+\angle EBC = 90^{\circ}。$ $\because \angle K=\angle EBC，$$\therefore \angle KCD=\angle ACB。$ $\therefore \overset{\frown}{DK}=\overset{\frown}{AB}。$ $\therefore DK = AB = 2。$ 在$Rt\triangle KCD$中，$\because DC = 4，$$DK = 2，$ $\therefore KC=\sqrt{2^{2}+4^{2}}=2\sqrt{5}。$ $\therefore \odot O$的半径为$\sqrt{5}。$ ;