【答案】
分析:(1)如圖1,連接OE,OD,由題意得,DE=DA=10,利用(SSS)判定△AOD≌△EOD,從可得∠OED=∠OAD=90°即可.
(2)當(dāng)點(diǎn)E運(yùn)動(dòng)到與B點(diǎn)重合的位置時(shí),如圖2,DE為正方形ABCD的對(duì)角線,所以此時(shí)DE最長(zhǎng),利用勾股定理求得DE,證明當(dāng)點(diǎn)E運(yùn)動(dòng)到線段OD與半圓O的交點(diǎn)處時(shí),DE最短.然后求得DE=OD-OE即可.
(3)當(dāng)點(diǎn)E與點(diǎn)A重合時(shí),DE=DA=10,此時(shí),直線DE的解析式為y=10;如圖4,當(dāng)點(diǎn)E與點(diǎn)A不重合時(shí),過(guò)點(diǎn)E作GH⊥x軸,分別交AD,x軸于點(diǎn)G,H,連接OE.則四邊形AFEG是矩形,且DE為圓O的切線,求證△OFE∽△DGE,利用其對(duì)應(yīng)邊成比例,設(shè)E(m,n),則有:EF=m,OF=OB-FB=5-n求得即可.
解答:證明:(1)如圖1,連接OE,OD,由題意得,
DE=DA=10,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/0.png)
,OD為公共邊
∴△AOD≌△EOD(SSS)
∴∠OED=∠OAD=90°
∴OE⊥DE,
∴DE與圓O相切.
(2)當(dāng)點(diǎn)E運(yùn)動(dòng)到與B點(diǎn)重合的位置時(shí),如圖2,DE為正方形ABCD的對(duì)角線,所以此時(shí)DE最長(zhǎng),
有:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/1.png)
,
當(dāng)點(diǎn)E運(yùn)動(dòng)到線段OD與半圓O的交點(diǎn)處時(shí),DE最短.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/images2.png)
證明如下:
在半圓O上任取一個(gè)不與點(diǎn)E重合的點(diǎn)E′,連接OE′,DE′.如圖3,
在△ODE′中,∵OE′+DE′>OD即:OE′+DE′>OE+DE,
∵OE′=OE,
∴DE′>DE
∵點(diǎn)E′是任意一個(gè)不與點(diǎn)E重合的點(diǎn),∴此時(shí)DE最短.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/2.png)
,
(3)當(dāng)點(diǎn)E與點(diǎn)A重合時(shí),DE=DA=10,此時(shí),直線DE的解析式為y=10;如圖4,
當(dāng)點(diǎn)E與點(diǎn)A不重合時(shí),過(guò)點(diǎn)E作GH⊥x軸,分別交AD,x軸于點(diǎn)G,H,連接OE.
則四邊形AFEG是矩形,
連接OD,
∵AD=DE,OA=OE,OD=OD,
∴△AOD≌△EOD,
∴∠OED=90°,
∴DE為圓O的切線
∴∠FEG=∠OED=90°
∴∠FEO=∠GED,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/images4.png)
又∵∠OFE=∠DGE=90°
∴△OFE∽△DGE
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/3.png)
,
設(shè)E(m,n),則有:EF=m,OF=OB-FB=5-n
得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/4.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/5.png)
,即:E(4,2)
又直線DE過(guò)點(diǎn)D(10,10),設(shè)直線DE解析式為y=kx+b,則有:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/6.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/7.png)
,即:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/8.png)
∴當(dāng)DE=10時(shí),直線DE的解析式為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/9.png)
;
以下兩種解法涉及高中知識(shí),僅供參考:
另解2:
(1)當(dāng)點(diǎn)E與點(diǎn)A重合時(shí),DE=DA=10,此時(shí),直線DE的解析式為y=10;
(2)當(dāng)點(diǎn)E與點(diǎn)A不重合時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/10.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/11.png)
設(shè)直線
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/12.png)
且經(jīng)過(guò)點(diǎn)(10,10),代入求得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/13.png)
所以直線DE的解析式為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/14.png)
;
另解3:
依題意得:點(diǎn)O的坐標(biāo)為(0,5),設(shè)直線DE的解析式為y=kx+b
由點(diǎn)到直線的距離公式得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/15.png)
,即(b-5)
2=25(k
2+1)①
直線DE過(guò)點(diǎn)D(10,10),得10=10k+b②
由①②解得:75k
2-100k=0,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/16.png)
所以直線DE的解析式為:為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155053029853276/SYS201310221550530298532023_DA/17.png)
.
點(diǎn)評(píng):此題涉及到的知識(shí)點(diǎn)較多,有相似三角形的判定與性質(zhì),待定系數(shù)法求一次函數(shù)解析式,勾股定理,切線的判定與性質(zhì),綜合性很強(qiáng),是一道很典型的題目.