【答案】
分析:(1)連接BC,由已知得∠ACB=2∠AOB=60°,AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/0.png)
AO=5,根據(jù)弧長(zhǎng)公式求解;
(2)連接OD,由垂直平分線的性質(zhì)得OD=OA=10,又DE=8,在Rt△ODE中,由勾股定理求OE,依題意證明△OEF∽△DEA,利用相似比求EF;
(3)存在.當(dāng)以點(diǎn)E、C、F為頂點(diǎn)的三角形與△AOB相似時(shí),分為①當(dāng)交點(diǎn)E在O,C之間時(shí),由以點(diǎn)E、C、F為頂點(diǎn)的三角形與△AOB相似,有∠ECF=∠BOA或∠ECF=∠OAB,②當(dāng)交點(diǎn)E在點(diǎn)C的右側(cè)時(shí),要使△ECF與△BAO相似,只能使∠ECF=∠BAO,③當(dāng)交點(diǎn)E在點(diǎn)O的左側(cè)時(shí),要使△ECF與△BAO相似,只能使∠ECF=∠BAO,三種情況,分別求E點(diǎn)坐標(biāo).
解答:解:(1)連接BC,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/images1.png)
∵A(10,0),∴OA=10,CA=5,
∵∠AOB=30°,
∴∠ACB=2∠AOB=60°,
∴弧AB的長(zhǎng)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/1.png)
;(4分)
(2)①若D在第一象限,
連接OD,
∵OA是⊙C直徑,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/images3.png)
∴∠OBA=90°,
又∵AB=BD,
∴OB是AD的垂直平分線,
∴OD=OA=10,
在Rt△ODE中,
OE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/3.png)
,
∴AE=AO-OE=10-6=4,
由∠AOB=∠ADE=90°-∠OAB,∠OEF=∠DEA,
得△OEF∽△DEA,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/4.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/5.png)
,
∴EF=3;(4分)
②若D在第二象限,
連接OD,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/images8.png)
∵OA是⊙C直徑,
∴∠OBA=90°,
又∵AB=BD,
∴OB是AD的垂直平分線,
∴OD=OA=10,
在Rt△ODE中,
OE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/7.png)
,
∴AE=AO+OE=10+6=16,
由∠AOB=∠ADE=90°-∠OAB,∠OEF=∠DEA,
得△OEF∽△DEA,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/8.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/10.png)
,
∴EF=12;
∴EF=3或12;
(3)設(shè)OE=x,
①當(dāng)交點(diǎn)E在O,C之間時(shí),由以點(diǎn)E、C、F為頂點(diǎn)的三角
形與△AOB相似,有∠ECF=∠BOA或∠ECF=∠OAB,
當(dāng)∠ECF=∠BOA時(shí),此時(shí)△OCF為等腰三角形,點(diǎn)E為OC
中點(diǎn),即OE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/11.png)
,
∴E
1(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/12.png)
,0);
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/images16.png)
當(dāng)∠ECF=∠OAB時(shí),有CE=5-x,AE=10-x,
∴CF∥AB,有CF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/13.png)
,
∵△ECF∽△EAD,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/14.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/15.png)
,解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/16.png)
,
∴E
2(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/17.png)
,0);
②當(dāng)交點(diǎn)E在點(diǎn)C的右側(cè)時(shí),
∵∠ECF>∠BOA,
∴要使△ECF與△BAO相似,只能使∠ECF=∠BAO,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/images22.png)
連接BE,
∵BE為Rt△ADE斜邊上的中線,
∴BE=AB=BD,
∴∠BEA=∠BAO,
∴∠BEA=∠ECF,
∴CF∥BE,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/18.png)
,
∵∠ECF=∠BAO,∠FEC=∠DEA=90°,
∴△CEF∽△AED,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/19.png)
,
而AD=2BE,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/20.png)
,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/21.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/22.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/23.png)
<0(舍去),
∴E
3(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/24.png)
,0);
③當(dāng)交點(diǎn)E在點(diǎn)O的左側(cè)時(shí),
∵∠BOA=∠EOF>∠ECF.
∴要使△ECF與△BAO相似,只能使∠ECF=∠BAO
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/images30.png)
連接BE,得BE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/25.png)
=AB,∠BEA=∠BAO
∴∠ECF=∠BEA,
∴CF∥BE,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/26.png)
,
又∵∠ECF=∠BAO,∠FEC=∠DEA=90°,
∴△CEF∽△AED,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/27.png)
,
而AD=2BE,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/28.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/29.png)
,
解得x
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/30.png)
,x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/31.png)
,
∵點(diǎn)E在x軸負(fù)半軸上,
∴E
4(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/32.png)
,0),
綜上所述:存在以點(diǎn)E、C、F為頂點(diǎn)的三角形與△AOB相似,
此時(shí)點(diǎn)E坐標(biāo)為:E
1(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/33.png)
,0)、E
2(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/34.png)
,0)、E
3(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/35.png)
,0)、E
4(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202757938605645/SYS201311032027579386056028_DA/36.png)
,0).(4分)
點(diǎn)評(píng):本題考查了相似三角形的判定與性質(zhì),勾股定理的運(yùn)用,圓周角定理,弧長(zhǎng)公式的運(yùn)用.關(guān)鍵是理解題意,根據(jù)基本條件,圖形的性質(zhì),分類求解.