答案：1. A 连接OA(图略),
∵半径$OC⊥AB,AB=8,\therefore AD=\frac {1}{2}AB=\frac {1}{2}×8=4,\because OA=OC=5,\therefore $在$Rt△OAD$中,$OD=\sqrt {OA^{2}-AD^{2}}=3$.故选A.

答案：
2. B 【解法一】如图,连接$OC,\because \widehat {AC}=\widehat {BC}$,AB为$\odot O$的直径,$\therefore ∠AOC=∠BOC=\frac {1}{2}∠AOB=90^{\circ },\therefore ∠D=\frac {1}{2}∠AOC=45^{\circ }$.故选B.

【解法二】如图,连接$AC,BC,\because \widehat {AC}=\widehat {BC}$,AB为$\odot O$的直径,$\therefore AC=BC,∠ACB=90^{\circ },\therefore ∠B=∠CAB=45^{\circ }$.$\because ∠D$和$∠B$都是$\widehat {AC}$所对的圆周角,$\therefore ∠D=∠B=45^{\circ }$.故选B.

答案：3. C 由圆内接四边形的性质可知,$∠ADC=180^{\circ }-∠ABC=180^{\circ }-70^{\circ }=110^{\circ },\because \widehat {AB}=\widehat {BC},\therefore ∠ADB=∠BDC=\frac {1}{2}∠ADC=55^{\circ }$.故选C.

答案：
4. C 如图,作点F关于AB的对称点$F'$,连接$EF',AC,OF$,则$EF'$与AB的交点就是所求的点P,此时$PE+PF=EF'$.

$\because F$为$\widehat {BC}$的中点,$\therefore ∠CAF=∠BAF$,$\because AE⊥OD,\therefore ∠AEC=∠AEO=90^{\circ }$,又$\because AE=AE,\therefore △AEC≌ △AEO(ASA),\therefore AC=OA$,$\because OA=OC,\therefore AC=OA=OC,\therefore △OAC$为等边三角形,$\therefore ∠AOC=∠OAC=60^{\circ },\because AO⊥AD,\therefore ∠DAO=90^{\circ }$,$\therefore ∠D=30^{\circ },\therefore OA=\frac {1}{2}OD$.$\because CD=4,\therefore OA=\frac {1}{2}(OA+4)$,解得$OA=4$,$\therefore AE=OA· sin∠AOC=4×sin60^{\circ }=2\sqrt {3}$,$\because OE⊥AF,\therefore AF=2AE=4\sqrt {3}$,连接$AF',\because $点F与点$F'$关于AB对称,$\therefore \widehat {AF}=\widehat {AF'},\widehat {BF}=\widehat {BF'},\therefore AF=AF',∠BAF=∠BAF'$,$\because ∠OAC=60^{\circ },∠CAF=∠BAF$,$\therefore ∠CAF=∠BAF=∠BAF'=30^{\circ }$,$\therefore ∠FAF'=60^{\circ },\therefore △AFF'$为等边三角形,$\therefore EF'⊥AF(E,P,F'$三点共线),$\therefore EF'=AF'· sin60^{\circ }=4\sqrt {3}×\frac {\sqrt {3}}{2}=6$,$\therefore PE+PF$的最小值为6,故选C.

答案：5. 答案 6
解析 $\because OC⊥AB,\therefore AB=2AC=2×3=6,\therefore OA=AB=6$.

答案：6. 答案 40
解析 $\because ∠BAC=50^{\circ },\therefore ∠BOC=2∠BAC=100^{\circ }$,$\because OB=OC,\therefore ∠OBC=∠OCB=\frac {180^{\circ }-∠BOC}{2}=40^{\circ }$.

答案：
7. 答案 $6\sqrt {3}$
解析 如图,作直径DE,连接BE,由圆周角定理的推论得$∠A=∠E,∠EBD=90^{\circ }$.$\because $四边形ABCD内接于$\odot O,\therefore ∠A+∠BCD=180^{\circ },\because ∠BCD=120^{\circ },\therefore ∠A=60^{\circ },\therefore ∠E=60^{\circ }$.$\because \odot O$的半径为6,$\therefore DE=12$,$\therefore BD=DE· sinE=12×sin60^{\circ }=12×\frac {\sqrt {3}}{2}=6\sqrt {3}$.

答案：8. 解析 (1)证明:$\because \widehat {BC}=\widehat {CD},\therefore ∠BOC=∠DOC$,在$△BOC$和$△DOC$中,$\{\begin{array}{l} OB=OD,\\ ∠BOC=∠DOC,\\ OC=OC,\end{array} $$\therefore △BOC≌ △DOC(SAS)$.
(2)$\because OC=OB,\therefore ∠ABC=∠OCB=65^{\circ }$,$\therefore ∠COB=180^{\circ }-∠ABC-∠OCB=50^{\circ }$,$\therefore ∠DOC=∠BOC=50^{\circ }$,$\therefore ∠AOD=180^{\circ }-∠DOC-∠BOC=80^{\circ }$,$\therefore ∠ABD=\frac {1}{2}∠AOD=40^{\circ }$.