已知△ABC,AC=BC,CD⊥AB于點D,點F在BD上,連接CF,AM⊥CF于點M,AM交CD于點E.
(1)如圖1,當(dāng)∠ACB=90°時,求證:DE=DF;
(2)如圖2,當(dāng)∠ACB=60°時,DE與DF的數(shù)量關(guān)系是______
【答案】
分析:(1)此題需先根據(jù)已知條件得出AD=CD,∠DCF=∠FAM,∠ADE=∠FDC,再根據(jù)AAS證出△ADE≌△CDF,即可得出DE=DF;
(2)根據(jù)∠ACB=60°,得出△ABC是等邊三角形,從而得出∠ACB=30°,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/0.png)
=Ctan30°=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/1.png)
,再根據(jù)△ADE∽△CDF,得出
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/3.png)
的值,即可得出DE與DF的數(shù)量關(guān)系;
(3根據(jù)已知條件得出EC的值,再設(shè)DE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/4.png)
x,則AD=4x,CD=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/5.png)
x,CE=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/6.png)
x,求出x的值,根據(jù)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/9.png)
,得出EP∥AB,從而證出△EPT為等邊三角形,求出EG的值,從而得出EP=ET=3,即可求出線段GT的長.
解答:解:(1)∵AC=BC,∠ACB=90°,
∴∠BAC=45°,
∵CD⊥AB,
∴∠ACD=45°,∠DFC+∠DCF=90°,
∴AD=CD,
∵AM⊥CF,
∴∠DFC+∠FAM=90°,
∴∠DCF=∠FAM,
∴△ADE≌△CDF,
∴DE=DF;
(2)∵∠ACB=60°,
∴△ABC是等邊三角形,
∴∠ACB=30°,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/10.png)
=Ctan30°=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/11.png)
,
∵∠ADE=∠FDC,∠DAE=∠DCF,
∴△ADE∽△CDF,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/14.png)
,
∴DF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/15.png)
DE;
(3)∵tan∠EAF=tan∠ECM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/16.png)
,EM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/17.png)
,
∴EC=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/18.png)
,
設(shè)DE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/19.png)
x,則AD=4x,CD=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/20.png)
x,CE=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/21.png)
x,
∴x=1,
∴AD=4,DE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/22.png)
,AE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/23.png)
,AB=8,
∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/24.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/25.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/26.png)
,
∴EP∥AB,
∵DF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/27.png)
DE,
∴∠PET=60°,
∴△EPT為等邊三角形,
∴EG∥AC,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/28.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/29.png)
,
∴EG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/30.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/31.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/32.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/33.png)
,
∴EP=ET=3,
∴GT=ET-GT=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163929145996608/SYS201310221639291459966027_DA/34.png)
;
點評:此題考查了等腰直角三角形,全等三角形和相似三角形的判定與性質(zhì),銳角三角函數(shù)值,平行線分線段成比例等知識點;是一道綜合題,解題時要注意有關(guān)知識的綜合應(yīng)用.