分班考试每日一题(10)
【题1】已知:在?ABC中,AD为BAC?的平分线,以C为圆心,CD为半径的半圆交BC的延长线于
点E,交AD于点F,交AE于点M,且:4:3BCAEFEFD????,.
⑴求证:AFDF?
⑵求?AED的余弦值;
⑶如果10BD?,求?ABC的面积.
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E
M
F
DCB
A
NE
M
F
DCB
A
NE
MF
DCB
A
【答案】⑴∵AD平分?BAC
∴BADDAC???,∵BCAE???,∴BADBDACCAE???????
∵ADEBADB?????,∴ADEDAE???,∴EAED?
∵DE是半圆C的直径
∴90DFE???
∴AFDF?
⑵过A点作ANBE?于N
在RtDFE?中,
∵:4:3FEFD?,∴可设4FEx?,则3FDx?
由勾股定理,得DEx?5
∴53AEDExAFFDx????,
∵11
22ADESADEFDEAN?????
,∴ADEFDEAN???,
∴(33)45xxxxAN????,∴24
5ANx?
∴由勾股定理,得ENx?75
∴775cos525xENAEDAEx????
⑶解法一:
过A点作ANBE?于N
由cos??AED725
得sin??AED2425,∴2424
255ANAEx??
在CAE?和ABE?中
∵CAEBAECBEA??????,
∴CAEABE??∽,∴AECE
BEAE?
∴2AEBECE??,∴25(5)(105)
2xxx???
解得x?27分
∴2448
55ANx??
,∴510215
2BCBDDC??????
∴11481572
225ABCSBCAN???????
解法二:
在CAE?和BE??中
∵CAEBAECBEA??????,,∴AEAE??∽,∴AECE
BEAE?
∴2AEBECE??,∴25(5)(105)
2xxx???
解得x?2
∴2448
55ANx??
BCBDDC??????1052215
∴11481572
225ABCSBCAN???????
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