【答案】
分析:(1)若⊙O與EC相切,且切點(diǎn)為D,可過D作EC的垂線,此垂線與AC的交點(diǎn)即為所求的O點(diǎn).
(2)由(1)知OD⊥EC,則∠ODA、∠E同為∠ADE的余角,因此∠E=∠ODA=∠OAD,而AD∥BC,可得∠OAD=∠ACB,等量代換后即可證得∠E=∠ACB.
(3)由(2)證得∠E=∠ACB,即tan∠E=tan∠DAC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/0.png)
,那么BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/1.png)
AB;由于AD∥BC,易證得△EAD∽△EBC,可用AB表示出AE、BC的長,根據(jù)相似三角形所得比例線段即可求出AB的長,進(jìn)而可得到BC的值.
解答:(1)解:(O即為AD中垂線與AC的交點(diǎn))或(過D點(diǎn)作EC的垂線與AC的交點(diǎn)等).
能見作圖痕跡,作圖基本準(zhǔn)確即可,漏標(biāo)O可不扣分(2分)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/images2.png)
(2)證明:連接OD.∵AD∥BC,∠B=90°,∴∠EAD=90°.
∴∠E+∠EDA=90°,即∠E=90°-∠EDA.
又∵圓O與EC相切于D點(diǎn),∴OD⊥EC.
∴∠EDA+∠ODA=90°,即∠ODA=90°-∠EDA.
∴∠E=∠ODA;(3分)
(說明:任得出一個角相等都評1分)
又∵OD=OA,∴∠DAC=∠ODA,∴∠DAC=∠E. (4分)
∵AD∥BC,∴∠DAC=∠ACB,∴∠E=∠ACB. (5分)
(3)解:Rt△DEA中,tanE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/2.png)
,又tanE=tan∠DAC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/3.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/images5.png)
∵AD=1,∴EA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/4.png)
. (6分)
Rt△ABC中,tan∠ACB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/5.png)
,
又∠DAC=∠ACB,∴tan∠ACB=tan∠DAC.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/7.png)
,∴可設(shè)AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/8.png)
x,BC=2x,
∵AD∥BC,∴Rt△EAD∽Rt△EBC. (7分)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/10.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160643622417072/SYS201310221606436224170024_DA/12.png)
.
∴x=1,
∴BC=2x=2. (8分)
點(diǎn)評:此題主要考查了切線的性質(zhì)、直角三角形的性質(zhì)、相似三角形的判斷和性質(zhì)等重要知識,能夠準(zhǔn)確的判斷出O點(diǎn)的位置,是解答此題的關(guān)鍵.