已知?ABCD的周長為28,自頂點A作AE⊥DC于點E,AF⊥BC于點F.若AE=3,AF=4,則CE-CF= .
【答案】
分析:首先可證得△ADE∽△ABF,又由四邊形ABCD是平行四邊形,即可求得AB與AD的長,然后根據(jù)勾股定理即可求得DE與BF的長,繼而求得答案.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/images0.png)
解:如圖1:∵AE⊥DC,AF⊥BC,
∴∠AED=∠AFB=90°,
∵四邊形ABCD是平行四邊形,
∴∠ADC=∠CBA,AB=CD,AD=BC,
∴△ADE∽△ABF,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/0.png)
,
∵AD+CD+BC+AB=28,
即AD+AB=14,
∴AD=6,AB=8,
∴DE=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/1.png)
,BF=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/2.png)
,
∴EC=CD-DE=8-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/3.png)
,CF=BF-BC=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/4.png)
-6,
∴CE-CF=(8-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/5.png)
)-(4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/6.png)
-6)=14-7
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/7.png)
;
如圖2:∵AE⊥DC,AF⊥BC,
∴∠AED=∠AFB=90°,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/images9.png)
∵四邊形ABCD是平行四邊形,
∴∠ADC=∠CBA,AB=CD,AD=BC,
∴∠ADE=∠ABF,
∴△ADE∽△ABF,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/8.png)
,
∵AD+CD+BC+AB=28,
即AD+AB=14,
∴AD=6,AB=8,
∴DE=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/9.png)
,BF=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/10.png)
,
∴EC=CD+DE=8+3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/11.png)
,CF=BC+BF=6+4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/12.png)
,
∴CE-CF=(8+3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/13.png)
)-(6+4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/14.png)
)=2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/15.png)
.
∴CE-CF=14-7
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/16.png)
或2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191817185956940/SYS201311011918171859569014_DA/17.png)
.
點評:本題主要考查的是平行四邊形的性質(zhì).解題時,還借用了勾股定理這一知識點.