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(2013•嘉定区一模)已知:点D是Rt△ABC的BC边的一个动点(如图),过点D作DE⊥AB,垂足为E,点F在AB边上(点F与点B不重合),且满足FE=BE,联结CF、DF.(1)当DF平分∠CFB时,求证:CFCB=BDFB
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(2013•嘉定区一模)已知:点D是Rt△ABC的BC边的一个动点(如图),过点D作DE⊥AB,垂足为E,点F在AB边上(点F与点B不重合),且满足FE=BE,联结CF、DF.
(1)当DF平分∠CFB时,求证:
=
:
(2)若AB=10,tanB=
.当DF⊥CF时,求BD的长.
![](http://hiphotos.baidu.com/zhidao/pic/item/e61190ef76c6a7efb9318f94fefaaf51f3de6668.jpg)
(1)当DF平分∠CFB时,求证:
CF |
CB |
BD |
FB |
(2)若AB=10,tanB=
3 |
4 |
![](http://hiphotos.baidu.com/zhidao/pic/item/e61190ef76c6a7efb9318f94fefaaf51f3de6668.jpg)
▼优质解答
答案和解析
(1)证明:∵DF平分∠CFB,
∴∠CFD=∠EFD,
∵DE⊥AB,FE=BE,
∴DF=BD,
∴∠EFD=∠DBF,
∵∠FCD=∠BCF,
∴△CFD∽△CBF,
∴
=
,
∵DF=BD,
∴
=
;![](http://hiphotos.baidu.com/zhidao/pic/item/fcfaaf51f3deb48ff17e866cf31f3a292df57868.jpg)
(2)∵AB=10,tanB=
,
∴AC=6,BC=8,
∵tanB=
.设DE=3x,则BE=4x,则BD=5x,CD=BC-BD=8-5x,
∵DE⊥AB,FE=BE,
∴DF=BD,
∴∠DFB=∠B,
∵DF⊥CF,
∴∠AFC+∠BFD=90°,
∵∠A+∠B=90°,
∴∠A=∠AFC,
∴AC=FC=6,
∴62+(5x)2=(8-5x)2,
解得:x=
,
故当DF⊥CF时,BD的长是
.
∴∠CFD=∠EFD,
∵DE⊥AB,FE=BE,
∴DF=BD,
∴∠EFD=∠DBF,
∵∠FCD=∠BCF,
∴△CFD∽△CBF,
∴
CF |
CB |
FD |
FB |
∵DF=BD,
∴
CF |
CB |
BD |
FB |
![](http://hiphotos.baidu.com/zhidao/pic/item/fcfaaf51f3deb48ff17e866cf31f3a292df57868.jpg)
(2)∵AB=10,tanB=
3 |
4 |
∴AC=6,BC=8,
∵tanB=
3 |
4 |
∵DE⊥AB,FE=BE,
∴DF=BD,
∴∠DFB=∠B,
∵DF⊥CF,
∴∠AFC+∠BFD=90°,
∵∠A+∠B=90°,
∴∠A=∠AFC,
∴AC=FC=6,
∴62+(5x)2=(8-5x)2,
解得:x=
7 |
20 |
故当DF⊥CF时,BD的长是
7 |
4 |
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