8. 如图，$△ ABC$内接于$\odot O$，$AB$为直径，$D$是$\overset{\frown}{AC}$上一点，且$∠ DAC=∠ DBA$，过点$D$作$DE⊥ AB$，垂足为$E$.
(1)求证:$BD$平分$∠ CBA$；
(2)连接$CD$，若$CD=5$，$BD=12$，求$\odot O$的半径.

8.(1)证明:$\because ∠DAC=∠DBC,∠DAC=∠DBA,$
$\therefore ∠DBA=∠CBD,\therefore BD$平分$∠CBA.$
(2)解:$\because ∠CBD=∠DBA,\therefore \overset{\frown}{AD}=\overset{\frown}{CD},\therefore AD=CD.$
$\because CD=5,\therefore AD=5.$
$\because AB$为直径,$\therefore ∠ADB=90^{\circ }.$
$\because BD=12,\therefore AB=\sqrt{AD^{2}+BD^{2}}=13,$
$\therefore \odot O$的半径为6.5.

9.如图,四边形ABCD是$\odot O$的内接四边形,连接AC,E为BC的延长线上一点,且CD平分∠ACE.
(1)如图①,若∠DCE=60°,求证:△ABD为等边三角形;
(2)如图②,若AB=10,BD=13,求$\odot O$的半径.

9.(1)证明:$\because CD$平分$∠ACE,\therefore ∠DCE=∠ACD.$
$\because ∠BAD+∠BCD=180^{\circ },∠BCD+∠DCE=180^{\circ },$
$\therefore ∠BAD=∠DCE=∠ACD.$
$\because ∠DCE=60^{\circ },\therefore ∠ABD=∠ACD=∠DCE=∠BAD=60^{\circ },$
$\therefore ∠ADB=180^{\circ }-60^{\circ }×2=60^{\circ }=∠ABD=∠BAD,$
$\therefore AB=AD=BD,\therefore △ABD$是等边三角形.
(2)解:如答图,过点$D$作$DG⊥AB$于点$G$,连接$OB,OA.$

由(1)知,$∠BAD=∠DCE=∠ACD=∠ABD,\therefore DB=DA.$
$\because AB=10,BD=13,\therefore BG=AG=5,$
$\therefore DG=\sqrt{13^{2}-5^{2}}=12,DG$垂直平分$AB.$
$\because OB=OA,\therefore$圆心$O$在$AB$的垂直平分线$DG$上,
$\therefore OG⊥AB$. 设$\odot O$的半径为$r$,在$\mathrm{Rt}△OBG$中,由勾股定理,得$r^{2}=(12-r)^{2}+5^{2},$
解得$r=\frac{169}{24},\therefore \odot O$的半径为$\frac{169}{24}.$

10. 如图，四边形ABCD是$\odot O$的内接四边形，且$AC⊥ BD$，垂足为E，AF是$\odot O$的直径。
(1)$∠ BAF$和$∠ CAD$相等吗？为什么？
(2)过圆心O作$OH⊥ AB$，垂足为H，若$OH=5$，求CD的长。

10.解:(1)$∠BAF$和$∠CAD$相等. 理由:如答图,连接$BF$.
$\because AF$是$\odot O$的直径,$\therefore ∠F+∠BAF=90^{\circ }.$
$\because AC⊥BD,\therefore ∠CAD+∠BDA=90^{\circ }.$
$\because ∠F=∠BDA,\therefore ∠BAF=∠CAD.$
(2)$\because OH⊥AB,\therefore AH=BH.$
$\because OA=OF,\therefore BF=2OH=10.$
$\because ∠BAF=∠CAD,\therefore CD=BF=10.$