【答案】
分析:(1)在Rt△OCD中,根據(jù)勾股定理易求OC=CD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/0.png)
.
(2)根據(jù)Rt△OAB的面積是
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/1.png)
可求出B點(diǎn)的坐標(biāo),因?yàn)锽D
2=AC
2+(AB-CD)
2,所以把B點(diǎn)的坐標(biāo)代入可得BD長(zhǎng),即可表示成關(guān)于t的函數(shù)關(guān)系式.
(3)假設(shè)OB=BD,在Rt△OAB中,用t把OB表示出來(lái),根據(jù)題(2)中用t表示的BD.兩者相等,可得一二次函數(shù)表達(dá)式,用根的判別式判斷是否有解.
(4)兩種情況,先假設(shè)∠EBD=90°時(shí)(如圖2),此時(shí)F、E、M三點(diǎn)重合,根據(jù)已知條件此時(shí)四邊形BDCF為直角梯形,然后假設(shè)∠EDB=90°時(shí)(如圖3),根據(jù)已知條件,此時(shí)四邊形BDCF為平行四邊形,在Rt△OCD中,OB
2=OD
2+BD
2,用t把各線段表示出來(lái)代入,可求出BD=CD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/2.png)
,即此時(shí)四邊形BDCF為菱形.
解答:解:(1)D(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/3.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/4.png)
);(1分)
(2)由Rt△OAB的面積為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/5.png)
,得B(t,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/6.png)
),
∵BD
2=AC
2+(AB-CD)
2,
∴BD
2=(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/7.png)
-t)
2+(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/8.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/9.png)
)
2=t
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/10.png)
-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/11.png)
(t+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/12.png)
)+4①
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/13.png)
,
∴BD=|t+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/14.png)
②;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/images15.png)
(3)解法一:若OB=BD,則OB
2=BD
2.
在Rt△OAB中,OB
2=OA
2+AB
2=t
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/15.png)
.
由①得t
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/16.png)
.
解得:t+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/17.png)
,∴t
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/18.png)
t+1=0,
∵△=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/19.png)
-4=-2<0,∴此方程無(wú)解.
∴OB≠BD.
解法二:若OB=BD,則B點(diǎn)在OD的中垂線CM上.
∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/20.png)
,
∴直線CM的函數(shù)關(guān)系式為y=-x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/21.png)
,③
由Rt△OAB的面積為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/22.png)
.④
聯(lián)立③,④得:x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/23.png)
x+1=0,
∵△=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/24.png)
-4=-2<0,∴此方程無(wú)解,
∴OB≠BD.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/images26.png)
解法三:若OB=BD,則B點(diǎn)在OD的中垂線CM上,如圖1
過點(diǎn)B作BG⊥y軸于G,CM交y軸于H,
∵S
△OBG=S
△OAB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/25.png)
,
而S
△OMH=S
△MOC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/26.png)
,(5分)
顯然與S
△HMO與S
△OBG矛盾.
∴OB≠BD.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/images29.png)
(4)如果△BDE為直角三角形,因?yàn)椤螧ED=45°,
①當(dāng)∠EBD=90°時(shí),此時(shí)F,E,M三點(diǎn)重合,如圖2
∵BF⊥x軸,DC⊥x軸,∴BF∥DC.
∴此時(shí)四邊形BDCF為直角梯形.
②當(dāng)∠EDB=90°時(shí),如圖3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/images30.png)
∵CF⊥OD,
∴BD∥CF.
又AB⊥x軸,DC⊥x軸,
∴BF∥DC.
∴此時(shí)四邊形BDCF為平行四邊形.
下證平行四邊形BDCF為菱形:
解法一:在△BDO中,OB
2=OD
2+BD
2,
∴t
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/27.png)
,
∴t+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/28.png)
,
[方法①]t
2-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/29.png)
t+1=0,∵BD在OD上方
解得:t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/30.png)
-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/31.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/32.png)
+1或t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/33.png)
+1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/34.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/35.png)
-1(舍去).
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/36.png)
,
[方法②]由②得:BD=t+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/37.png)
,
此時(shí)BD=CD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/38.png)
,
∴此時(shí)四邊形BDCF為菱形(9分)
解法二:在等腰Rt△OAE與等腰Rt△EDB中
∵OA=AE=t,OE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/39.png)
t,則ED=BD=2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/40.png)
t,
∴AB=AE+BE=t+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/41.png)
(2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/42.png)
t)=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/43.png)
-t,
∴2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/44.png)
以下同解法一,
此時(shí)BD=CD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103100800657370475/SYS201311031008006573704026_DA/45.png)
,
∴此時(shí)四邊形BDCF為菱形.(9分)
點(diǎn)評(píng):此題考查了一次函數(shù)解析式的確定、根的判別式、三角形面積的求法、菱形的判定以及勾股定理的應(yīng)用等知識(shí),綜合性強(qiáng),難度較大.