三、解答题(共40分)
11. (10分)如图,在$△ ABC$中,以边$AC$上一点$O$为圆心、$OA$为半径作$\odot O$,与$AB$相切于点$A$,作$CD\bot BO$交$BO$的延长线于点$D$,且$∠ CBD=∠ DCO$,求证:$BC$是$\odot O$的切线.

证明：如图，过点$O$作$OE\perp BC$于点$E.$$\because CD\perp BO，$$\therefore\angle D = 90^{\circ}，$$\therefore\angle BCD+\angle CBD = 90^{\circ}，$$\angle COD+\angle DCO = 90^{\circ}.$$\because\angle CBD=\angle DCO，$$\therefore\angle BCD=\angle COD=\angle BOA.$又$\because AB$为$\odot O$的切线，$\therefore AC\perp AB，$$\therefore\angle BAC=\angle D = 90^{\circ}.$$\because\angle BCD=\angle BOA，$$\therefore\angle OBA=\angle OBC，$$\therefore OE = OA.$$\because OE\perp BC，$$OE$是$\odot O$的半径，$\therefore BC$是$\odot O$的切线. ;

12. (10分)已知$∠ O$及其一边上的两点$A$、$B$. 求作$\odot P$,使圆心$P$到$AB$两点的距离相等,且与$∠ O$的两条边相切.

解: 如图，$\odot P$即为所求.作法：①作线段$AB$的垂直平分线$MN$与$\angle O$的平分线$OC$相交于点$P，$$MN$与$AB$的交点为垂足$D；$②以点$P$为圆心、$PD$为半径画圆. ;

13. (10 分)如图，$△ ABC$ 内接于$\odot O$，$D$ 是$\overset{\frown}{BC}$上一点，经过点 $D$ 的$\odot O$ 的切线 $EF // BC$，分别交 $AB$、$AC$ 的延长线于点 $E$、$F$。
(1)求证：$AD$ 平分 $∠ BAC$。
(2)若$\odot O$ 的半径为 $12$，$∠ BAC=60°$，$BE=6\sqrt{2}$，求线段 $DE$ 的长。

证明: (1)如图，连接$OD.$$\because EF$是$\odot O$的切线，$\therefore OD\perp EF.$$\because EF// BC，$$\therefore OD\perp BC，$$\therefore\overset{\frown}{BD}=\overset{\frown}{CD}，$$\therefore\angle BAD=\angle CAD，$$\therefore AD$平分$\angle BAC.$ (2)如图，连接$OB$、$BD，$过点$B$作$BH\perp DE$于点$H.$$\because\angle BAC = 60^{\circ}，$$AD$平分$\angle BAC，$即$\angle BAD=\frac{1}{2}\angle BAC，$$\therefore\angle BOD = 2\angle BAD=\angle BAC = 60^{\circ}.$$\because OB = OD，$$\therefore\triangle OBD$为等边三角形，$\therefore\angle BDH = 30^{\circ}，$$BD = OD = 12.$在$Rt\triangle BDH$中，$BH=\frac{1}{2}BD = 6，$$DH = 6\sqrt{3}，$$\therefore EH=\sqrt{BE^{2}-BH^{2}} = 6，$$\therefore DE = EH + HD = 6 + 6\sqrt{3}.$ ;

14. (10 分)如图，P 为$\odot O$外一点，PA 为$\odot O$的切线，A 为切点，作直径 AB，过点 B 作 $BC // PO$ 交$\odot O$于点 C.
(1)求证：PC 为$\odot O$的切线.
(2)连接 AC 交 PO 于点 M，若$\odot O$的半径为 3，$PA=4$，求 BC 的长.

证明: (1)如图，连接$OC.$$\because PA$为$\odot O$的切线，$\therefore OA\perp PA，$$\therefore\angle PAO = 90^{\circ}.$$\because OB = OC，$$\therefore\angle OCB=\angle B.$$\because BC// PO，$$\therefore\angle COP=\angle OCB，$$\angle AOP=\angle B，$$\therefore\angle COP=\angle AOP.$在$\triangle COP$和$\triangle AOP$中，$\begin{cases}OC = OA\\\angle COP=\angle AOP\\OP = OP\end{cases}，$$\therefore\triangle COP\cong\triangle AOP(SAS)，$$\therefore\angle OCP=\angle OAP = 90^{\circ}，$$\therefore OC\perp PC.$$\because OC$为$\odot O$的半径，$\therefore PC$为$\odot O$的切线. (2)$\because\odot O$的半径为$3，$$\therefore OA = OB = 3.$$\because PA = 4，$$\therefore PO=\sqrt{PA^{2}+OA^{2}} = 5.$由 (1)知，$\triangle COP\cong\triangle AOP，$$\therefore PA = PC，$$\angle CPO=\angle APO，$$\therefore PM\perp AC，$$AM = MC.$$\because S_{\triangle PAO}=\frac{1}{2}PA\cdot AO=\frac{1}{2}PO\cdot AM，$$\therefore PA\cdot AO = PO\cdot AM，$$\therefore 4\times3 = 5AM，$$\therefore AM=\frac{12}{5}，$$\therefore OM=\sqrt{OA^{2}-AM^{2}}=\sqrt{3^{2}-(\frac{12}{5})^{2}}=\frac{9}{5}.$$\because AM = MC，$$OA = OB，$$\therefore OM$为$\triangle ABC$的中位线，$\therefore BC = 2OM=\frac{18}{5}.$ ;