【答案】
分析:(1)由相似三角形,列出比例關(guān)系式,即可證明;
(2)首先求出矩形EFPQ面積的表達(dá)式,然后利用二次函數(shù)求其最大面積;
(3)本問是運(yùn)動(dòng)型問題,要點(diǎn)是弄清矩形EFPQ的運(yùn)動(dòng)過程:
(I)當(dāng)0≤t≤2時(shí),如答圖①所示,此時(shí)重疊部分是一個(gè)矩形和一個(gè)梯形;
(II)當(dāng)2<t≤4時(shí),如答圖②所示,此時(shí)重疊部分是一個(gè)三角形.
解答:(1)證明:∵矩形EFPQ,
∴EF∥BC,∴△AHF∽△ADC,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/0.png)
,
∵EF∥BC,∴△AEF∽△ABC,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/1.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/2.png)
.
(2)解:∵∠B=45°,∴BD=AD=4,∴CD=BC-BD=5-4=1.
∵EF∥BC,∴△AEH∽△ABD,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/3.png)
,
∵EF∥BC,∴△AFH∽△ACD,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/4.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/5.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/6.png)
,∴EH=4HF,
已知EF=x,則EH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/7.png)
x.
∵∠B=45°,∴EQ=BQ=BD-QD=BD-EH=4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/8.png)
x.
S
矩形EFPQ=EF•EQ=x•(4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/9.png)
x)=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/10.png)
x
2+4x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/11.png)
(x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/12.png)
)
2+5,
∴當(dāng)x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/13.png)
時(shí),矩形EFPQ的面積最大,最大面積為5.
(3)解:由(2)可知,當(dāng)矩形EFPQ的面積最大時(shí),矩形的長為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/14.png)
,寬為4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/15.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/16.png)
=2.
在矩形EFPQ沿射線AD的運(yùn)動(dòng)過程中:
(I)當(dāng)0≤t≤2時(shí),如答圖①所示.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/images17.png)
設(shè)矩形與AB、AC分別交于點(diǎn)K、N,與AD分別交于點(diǎn)H
1,D
1.
此時(shí)DD
1=t,H
1D
1=2,
∴HD
1=HD-DD
1=2-t,HH
1=H
1D
1-HD
1=t,AH
1=AH-HH
1=2-t,.
∵KN∥EF,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/17.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/18.png)
,得KN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/19.png)
(2-t).
S=S
梯形KNFE+S
矩形EFP1Q1
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/20.png)
(KN+EF)•HH
1+EF•EQ
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/21.png)
[
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/22.png)
(2-t)+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/23.png)
]×t+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/24.png)
(2-t)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/25.png)
t
2+5;
(II)當(dāng)2<t≤4時(shí),如答圖②所示.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/images27.png)
設(shè)矩形與AB、AC分別交于點(diǎn)K、N,與AD交于點(diǎn)D
2.
此時(shí)DD
2=t,AD
2=AD-DD
2=4-t,
∵KN∥EF,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/26.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/27.png)
,得KN=5-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/28.png)
t.
S=S
△AKN
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/29.png)
KN•AD
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/30.png)
(5-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/31.png)
t)(4-t)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/32.png)
t
2-5t+10.
綜上所述,S與t的函數(shù)關(guān)系式為:
S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193338456143253/SYS201311011933384561432024_DA/33.png)
.
點(diǎn)評(píng):本題是運(yùn)動(dòng)型相似三角形壓軸題,考查了相似三角形的判定與性質(zhì)、二次函數(shù)的表達(dá)式與最值、矩形、等腰直角三角形等多個(gè)知識(shí)點(diǎn),涉及考點(diǎn)較多,有一定的難度.難點(diǎn)在于第(3)問,弄清矩形的運(yùn)動(dòng)過程是解題的關(guān)鍵.