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重庆中考24题

发布时间:2014-03-12 15:59:42  

重庆中考第24题专题一

06)25. (10分)如图,在梯形ABCD中,AB//DC,∠BCD=90,且AB=1,BC=2,tan∠ADC=2.

⑴求证:DC=BC;

⑵E是梯形内的一点,F是梯形外的一点,且∠EDC=∠FBC,DE=BF,试判断△ECF的形状,并证明你的结论;

⑶在⑵的条件下,当BE:CE=1:2,∠BEC=135时,求sin∠BFE的值。

A

B

F

DC

07) 26.(10分)已知,如图:△ABC是等腰直角三角形,∠ABC=900,AB=10,D为△ABC外一点,边结AD、BD,过D作DH⊥AB,垂足为H,交AC于E。

(1)若△ABD是等边三角形,求DE的长;

(2)若BD=AB,且tan?HDB?

A??3,求DE的长。 4

DH

C

26 题图B

08) 26、(10分)已知:如图,在梯形ABCD中,AD∥BC,BC=DC,CF平分∠BCD,DF∥AB,BF的延长线交DC于点E。 求证:(1)△BFC≌△DFC;(2)AD=DE

26题图 1

09)24.已知:如图,在直角梯形ABCD中,AD∥BC,∠ABC=90o,DE⊥AC于点F,交BC于点G,交AB的延长线于点E,且AE=AC。 (1)求证:BG=FG;

(2)若AD=DC=2,求AB的长。

_

_ B _ C_ G

_ E

10) 24. 已知:如图,在直角梯形ABCD中,AD∥BC,∠ABC=90°.点E是DC的

中点,过点E作DC的垂线交AB于点P,交CB的延长线于点M.点F在线段ME上,且满足CF=AD,MF=MA.

(1)若∠MFC=120°,求证:AM=2MB;

1 (2)求证:∠MPB=90°- ∠FCM. 2

11)24、(2011?重庆)如图,梯形ABCD中,AD∥BC,∠DCB=45°,CD=2,BD⊥CD.过点C作CE⊥AB于E,交对角线BD于F,点G为BC中点,连接EG、AF.

(1)求EG的长;

(2)求证:CF=AB+AF.

2

12)24.(2012重庆)已知:如图,在菱形ABCD中,F为边BC的中点,DF与对角线AC交于点M,过M作ME⊥CD于点E,∠1=∠2.

(1)若CE=1,求BC的长;

(2)求证:AM=DF+ME.

练习:

24.如图,梯形ABCD中,AD∥BC,∠ABC=90°,AD=AB,DF⊥BC于F,连接AF,P为AF上一点,连接DP、CP,且DP⊥CP,CP交DF于G,CP的延长线交AB于E.

(1)若CD?DP的长;

(2)求证:BC=AD+AE;

PFCEAD

B

24.如图所示,在直角梯形ABCD中,?BCD?90?,AD//BC,CD?BC,E是CD上一点,

BE?AC.

(1)求证:AD?EC

(2)当点E在CD上什么位置时,AB?BE成立?并说明理由.

A D E 3

24. 如图,在矩形ABCD中,E、F分别是AB、CD上的点,AE=CF,连接EF、BF,EF与对角线AC交于点O,且BE=BF,∠BEF=2∠BAC。

(1)求证:OE=OF

(2)若BC=23,求AB的长。

作业:1、如图,AC是正方形ABCD的对角线,点O是AC的中点,点Q是AB上一点,连接CQ,DP⊥CQ于点E,交BC于点P,连接OP,OQ;

求证:

(1)△BCQ≌△CDP;

(2)OP=OQ.

2、如图,在正方形ABCD中,E、F分别为BC、AB上两点,且BE=BF,过点B作AE的垂线交AC于点G,过点G作CF的垂线交BC于点H延长线段AE、GH交于点M.

(1)求证:∠BFC=∠BEA; (2)求证:AM=BG+GM.

4

参考答案:

06,25.(1)过A作DC的垂线AM交DC于M,

则AM=BC=2. (1分)

又tan∠ADC=2,所以DM?2

2?1.(2分)

因为MC=AB=1,所以DC=DM+MC=2即DC=BC. (3分)

(2)等腰直角三角形. (4分)

证明:因为DE?DF,?EDC??FBC,DC?BC.

所以,△DEC≌△BFC(5分)

所以,CE?CF,?ECD??BCF.

所以,?ECF??BCF??BCE??ECD??BCE??BCD?90? 即△ECF是等腰直角三角形. (6分)

(3)设BE?k,则CE?CF?

2k,所以EF?.(7分)

因为?BEC?135?,又?CEF?45?,所以?BEF?90?.(8分)

所以BF??3k(9分) 所以sin?BFE?k

3k?1

3.(10分)

07、26.(1)∵△ABD是等边三角形,AB=10,∴∠ADB=600,AD=AB=10

∵DH⊥AB ∴AH=1

2AB=5

∴DH=AD2?AH2?2?52?5

∵△ABC是等腰直角三角形 ∴∠CAB=450

5

∴∠AEH=450 ∴EH=AH=5

∴DE=DH-EH=5?5

3 4

∴可设BH=3k,则DH=4k,DB=5k

∵BD=AB=10 ∴5k?10 解得:k?2 (2)∵DH⊥AB且tan?HDB?

∴DH=8,BH=6,AH=4

又∵EH=AH=4

∴DE=DH-EH=4

08、26.证明:(1)?CF平分?BCD,??BCF??DCF. ····························· (1分) 在△BFC和△DFC中,

?BC?DC,? ········································································································ (3分) ??BCF??DCF,

?FC?FC.?

?△BFC≌△DFC. ·································································································· (4分)

(2)连结BD. ············································································································· (5分) ?△BFC≌△DFC,

?BF?DF,

??FBD??FDB. ······································ (6分)

?DF∥AB,??ABD??FDB.

??ABD??FBD. ······································ (7分)

?AD∥BC,??BDA??DBC.

?BC?DC,??DBC??BDC.

??BDA??BDC. ········································ (8分) 26题图 又BD是公共边,?△BAD≌△BED. ······· (9分)

?AD?DE. ············································································································· (10分) 09、24 (1) DE⊥AC => ∠AFE=90 o => ∠EAC + ∠AEF = ∠EAC + ∠ACB = 90 o

=>∠EAC = ∠ACB

AB = AE, ∠EAC=∠CAE

△EAF 和△CAB全等。=> AB=AF

连接AG,直角三角形一条边和斜边相等 则△ABG和△AFG全等

=> BG=FG

(2) AD = DC 知△ADC为等腰三角形,DE⊥AC知 DF为AC的中垂线

AF = FC 又 AF=AB 得AB = 1/2 AC

∠ABC = 90 o 所以∠BAC=60 o ∠CAD=30 o

AB=AF = AD*COS60 o = 3

6

10

11解:①正确.因为AB=AD=AF,AG=AG,∠B=∠AFG=90°,∴△ABG≌△AFG; ②正确.因为:EF=DE=CD=2,设BG=FG=x,则CG=6﹣x.在直角△ECG中,根据勾股定理,得(6﹣x)+4=(x+2),解得x=3.所以BG=3=6﹣3=GC;

③正确.因为CG=BG=GF,所以△FGC是等腰三角形,∠GFC=∠GCF.又∠AGB=∠AGF,∠AGB+∠AGF=180°﹣∠FGC=∠GFC+∠GCF,

∴∠AGB=∠AGF=∠GFC=∠GCF,∴AG∥CF;

④错误.

过F作FH⊥DC,

∵BC⊥DH,

∴FH∥GC,

∴△EFH∽△EGC, ∴=, 222

EF=DE=2,GF=3,

∴EG=5,

7

∴==,

∴S△FGC=S△GCE﹣S△FEC=×3×4﹣×4×(×3)=≠3. 故选C.

12、(1)解:∵四边形ABCD是菱形, ∴AB∥CD,

∴∠1=∠ACD,

∵∠1=∠2,

∴∠ACD=∠2,

∴MC=MD,

∵ME⊥CD,

∴CD=2CE,

∵CE=1,

∴CD=2,

∴BC=CD=2;

(2)证明:如图,∵F为边BC的中点, ∴

BF=CF=BC,

∴CF=CE,

在菱形ABCD中,AC平分∠BCD, ∴∠ACB=∠ACD,

在△CEM和△CFM中, ∵,

∴△CEM≌△CFM(SAS),

∴ME=MF,

延长AB交DF于点G,

∵AB∥CD,

∴∠G=∠2,

∵∠1=∠2,

∴∠1=∠G,

∴AM=MG,

在△CDF和△BGF中,

8

∵, ∴△CDF≌△BGF(AAS),

∴GF=DF,

由图形可知,GM=GF+MF, ∴AM=DF+ME.

9

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