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【330819】菱形的判定

时间:2025-02-11 18:48:31 作者: 字数:15076字

湘教版8年级下册数学2.6.2菱形的判定同步练习

一、选择题(本大题共8小题)

1. 如图,要使ABCD成为菱形,则需添加的一个条件是(  )

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

AAC=AD BBA=BC C.∠ABC=90° D AC=BD

2. 如图,将△ABC沿BC方向平移得到△DCE,连接AD,下列条件能够判定四边形ACED为菱形的是(  )

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

AAB=BC BAC=BC C.∠B=60° D.∠ACB=60°

3. 如图,四边形ABCD的四边相等,且面积为120cm2,对角线AC=24cm,则四边形ABCD的周长为(  )

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

A52cm B40cm C39cm D26cm

4. 如图,点ABCD在同一条直线上,点EF分别在直线AD的两侧,且AE=DF,∠A=∠DAB=DC.若AD=10DC=3,∠EBD=60°,则BE为( )时,四边形BFCE是菱形.

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

A5 B4 C3 D6

5. 如图在Rt△ABC中,∠ACB=90°AC=4BC=3D为斜边AB上一点,以CDCB为边作平行四边形CDEB,当AD的值为( )时,平行四边形CDEB为菱形.

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

A14 B16 C18 D10

6. 如图,已知矩形ABCD的对角线长为8cmEFGH分别是ABBCCDDA的中点,则四边形EFGH的周长是( )cm

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

A14 B16 C18 D10

7. 如图,矩形ABCD的对角线ACBD相交于点OCEBDDEAC,若AC=4,则四边形CODE的周长(  )

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

A4 B6 C8 D10

8. 过矩形ABCD的对角线AC的中点OEFAC,交BC边于点E,交AD边于点F,分别连接AECF.若AB= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> ,∠DCF=30°,则EF的长为(  )

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

  A2 B3 C <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> D <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

二、填空题(本大题共6小题)

9. 如图,平行四边形ABCD的对角线ACBD相交于点O,请你添加一个适当的条件   使其成为菱形(只填一个即可).

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

10. 如图,四边形ABCD内有一点EAE=BE=DE=BC=DCAB=AD,若C=100°,则BAD的大小是

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

11. 如图,已知菱形ABCD的一个内角∠BAD=80°,对角线ACBD相交于点O,点EAB上,且BE=BO,则∠EOA=______.

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

12. 如图,在△ABC,点DBC的中点,点EF分别在线段AD及其延长线上,且DE=DF,给出下列条件:①BE⊥EC;②BF∥CE;③AB=AC,从中选择一个条件使四边形BECF是菱形,你认为这个条件是__________(填序号).

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>


13. 如图,四边形ABCD是轴对称图形,且直线AC是对称轴,AB∥CD,则下列结论:①AC⊥BD;②AD∥BC;③四边形ABCD是菱形;④△ABD≌△CDB.其中正确的是   (只填写序号)

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

14. 如图,在给定的一张平行四边形纸片上做一个菱形,甲、乙两人的作法如下:

甲:连接AC,做AC的垂直平分线MN分别交ADACBCMON,连接ANCM,则四边形ANCM是菱形.

乙:分别作∠A,∠B的平分线AEBF,分别交BCADEF,连接EF,则四边形ABEF是菱形.

根据两人的作法可判断正确的是

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

三、计算题(本大题共4小题)

15. 如图,在△ABC中,∠ACB=90°DE分别为ACAB的中点,BFCEDE的延长线于点F

1)求证:四边形ECBF是平行四边形;

2)当∠A=30°时,求证:四边形ECBF是菱形.

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>


16. 如图,在RtABC中,∠ACB=90°DAB的中点,且AECDCEAB

1)证明:四边形ADCE是菱形;

2)若∠B=60°BC=6,求菱形ADCE的高.(计算结果保留根号)

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>








17. 如图,△ABC≌△ABD,点E在边AB上,CEBD,连接DE.求证:

1)∠CEB=CBE

2)四边形BCED是菱形.

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>







18. 如图,AEBFAC平分∠BAE,且交BF于点CBD平分∠ABF,且交AE于点DACBD相交于点O,连接CD

1)求∠AOD的度数;

2)求证:四边形ABCD是菱形.

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>






19. 如图,在平行四边形ABCD中,EF分别为边ABCD的中点,BD是对角线.

1)求证:△ADE≌△CBF

2)若∠ADB是直角,则四边形BEDF是什么四边形?证明你的结论.

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>












参考答案:

一、选择题(本大题共8小题)

1. B

分析:利用邻边相等的平行四边形为菱形即可得证.

解:如图,要使ABCD成为菱形,则需添加的一个条件是BA=BC

故选B

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

2. B

分析:首先根据平移的性质得出AB <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> CD,得出四边形ABCD为平行四边形,进而利用菱形的判定得出答案.

解:∵将△ABC沿BC方向平移得到△DCE

AB <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> CD

四边形ABCD为平行四边形,

AC=BC时,

平行四边形ACED是菱形.

故选:B

3. A

分析:可定四边形ABCD为菱形,连ACBD相交于点O,则可求得BD的长,在RtAOB中,利用勾股定理可求得AB的长,从而可求得四边形ABCD的周长.

解:如图,连接ACBD相交于点O

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

四边形ABCD的四边相等,

四边形ABCD为菱形,

ACBDS四边形ABCD= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> AC•BD

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> ×24BD=120,解得BD=10cm

OA=12cmOB=5cm

RtAOB中,由勾股定理可得AB= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> =13cm),

四边形ABCD的周长=4×13=52cm),故选A

4. B

分析:(1)由AE=DF,∠A=DAB=DC,易证得△AEC≌△DFB,即可得BF=EC,∠ACE=∠DBF,且ECBF,即可判定四边形BFCE是平行四边形;

2)当四边形BFCE是菱形时,BE=CE,根据菱形的性质即可得到结果.

解:(1)证明:∵AB=DC

AC=DF

在△AEC和△DFB

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

∴△AEC≌△DFBSAS),

BF=EC,∠ACE=∠DBF

ECBF

四边形BFCE是平行四边形;

2)当四边形BFCE是菱形时,BE=CE

AD=10DC=3AB=CD=3

BC=10﹣3﹣3=4

∵∠EBD=60°

BE=BC=4

BE=4 时,四边形BFCE是菱形,

故答案为:4.故选B.

5. C

分析:首先根据勾股定理求得AB=5;然后利用菱形的对角线互相垂直平分、邻边相等推知OD=OBCD=CB;最后Rt△BOC中,根据勾股定理得,OB的值,则AD=AB-2OB

解:如图,连接CEAB于点O

Rt△ABC中,∠ACB=90°AC=4BC=3

AB= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> =5(勾股定理).

若平行四边形CDEB为菱形时,CE⊥BD,且OD=OBCD=CB

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> AB•OC= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> AC•BC

OC= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

Rt△BOC中,根据勾股定理得,OB= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> = <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> = <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>
∴AD=AB-2OB=
 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

故答案是: <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

6. B

分析:利用三角形的中位线定理;矩形的性质;菱形的判定及性质解答即可。

解:根据三角形的中位线定理和矩形对角线相等的性质可证得四边形EFGH是菱形,且EF=  <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> AC=4,所以菱形的周长等于16cm,故选B

7. C

分析:首先由CEBDDEAC,可证得四边形CODE是平行四边形,又由四边形ABCD是矩形,根据矩形的性质,易得OC=OD=2,即可判定四边形CODE是菱形,继而求得答案.

解:∵CEBDDEAC

四边形CODE是平行四边形,

四边形ABCD是矩形,

AC=BD=4OA=OCOB=OD

OD=OC= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> AC=2

四边形CODE是菱形,

四边形CODE的周长为:4OC=4×2=8

故选C

8. A

分析: 求出∠ACB=DAC,然后利用“角角边”证明△AOF和△COE全等,根据全等三角形对应边相等可得OE=OF,再根据对角线互相垂直平分的四边形是菱形得到四边形AECF是菱形,再求出∠ECF=60°,然后判断出△CEF是等边三角形,根据等边三角形的三条边都相等可得EF=CF,根据矩形的对边相等可得CD=AB,然后求出CF,从而得解.

解答: 解:∵矩形对边ADBC

∴∠ACB=DAC

OAC的中点,

AO=CO

在△AOF和△COE中,

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

∴△AOFCOEASA),

OE=OF

又∵EFAC

四边形AECF是菱形,

∵∠DCF=30°

∴∠ECF=90°30°=60°

∴△CEF是等边三角形,

EF=CF

AB= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

CD=AB= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

∵∠DCF=30°

CF= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> ÷ <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> =2

EF=2

故选A

二、填空题(本大题共6小题)

9. 分析:利用菱形的判定方法确定出适当的条件即可.

解:如图,平行四边形ABCD的对角线ACBD相交于点O,添加一个适当的条件为:ACBD或∠AOB=90°AB=BC使其成为菱形.

故答案为:ACBD或∠AOB=90°AB=BC

10. 分析:由题干BE=DE=BC=DC,可知四边形BECD为菱形,又C=100°,所以BED=100°CBE=CDE=80°.连接BD,易知AEBEDEABD的角平分线.再根据菱形的性质即可得出答案.

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

解:连接BD,并延长AEBD于点O

AE=BE=DE=BC=DCAB=AD四边形BCDE是菱形,

AEBEDEABD的角平分线.

AEOC四点共线,

∵∠C=100°∴∠BED=50°

∴∠BEO= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> BED=50°

∴∠ABE=25°

∴∠BAD=50°

11. 分析:因为AB=ADBAD=80°,可求ABD=50°;又BE=BO,所以BEO=BOE,根据三角形内角和定理求解.

解:∵ABCD是菱形,AB=AD∴∠ABD=ADB

∵∠BAD=80°∴∠ABD= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> ×180°-80°=50°

BE=BO

∴∠BEO=BOE= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> ×180°-50°=65°

故答案为:65

12.分析:根据菱形的判定方法进行验证得到答案。

解:∵BD=CDDE=DF

四边形BECF是平行四边形,

BEEC时,四边形BECF是矩形,不一定是菱形;

四边形BECF是平行四边形,则BFEC一定成立,故不一定是菱形;

AB=AC时,∵DBC的中点,

AFBC的中垂线,

BE=CE

平行四边形BECF是菱形.

13. 分析:根据轴对称图形的性质,结合菱形的判定方法以及全等三角形的判定方法分析得出答案.

解:因为l是四边形ABCD的对称轴,AB∥CD

AD=AB,∠1=∠2,∠1=∠4

则∠2=∠4

AD=DC

同理可得:AB=AD=BC=DC

所以四边形ABCD是菱形.

根据菱形的性质,可以得出以下结论:

所以①AC⊥BD,正确;

AD∥BC,正确;

四边形ABCD是菱形,正确;

在△ABD和△CDB

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

∴△ABD≌△CDBSSS),正确.

故答案为:①②③④.

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

14.分析:首先证明△AOM≌△CONASA),可得MO=NO,再根据对角线互相平分的四边形是平行四边形可判定判定四边形ANCM是平行四边形,再由ACMN,可根据对角线互相垂直的四边形是菱形判定出ANCM是菱形;四边形ABCD是平行四边形,可根据角平分线的定义和平行线的定义,求得AB=AF,所以四边形ABEF是菱形.

解:甲的作法正确;

四边形ABCD是平行四边形,

ADBC

∴∠DAC=ACN

MNAC的垂直平分线,

AO=CO

在△AOM和△CON <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

∴△AOM≌△CONASA),

MO=NO

四边形ANCM是平行四边形,

ACMN

四边形ANCM是菱形;

乙的作法正确;

ADBC

∴∠1=2,∠6=7

BF平分∠ABCAE平分∠BAD

∴∠2=3,∠5=6

∴∠1=3,∠5=7

AB=AFAB=BE

AF=BE

AFBE,且AF=BE

四边形ABEF是平行四边形,

AB=AF

平行四边形ABEF是菱形;

故选:C

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

三、计算题(本大题共4小题)

15. 分析:(1)利用平行四边形的判定证明即可;

2)利用菱形的判定证明即可.

证明:(1)∵DE分别为边ACAB的中点,

DEBC,即EFBC

又∵BFCE

四边形ECBF是平行四边形.

2)∵∠ACB=90°,∠A=30°EAB的中点,

CB= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> ABCE= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> AB

CB=CE

又由(1)知,四边形ECBF是平行四边形,

四边形ECBF是菱形.

16.分析:(1)先证明四边形ADCE是平行四边形,再证出一组邻边相等,即可得出结论;

2)过点DDFCE,垂足为点F;先证明△BCD是等边三角形,得出∠BDC=BCD=60°CD=BC=6,再由平行线的性质得出∠DCE=BDC=60°,在RtCDF中,由直角三角形的性质求出DF即可.

解答:(1)证明:∵AECDCEAB

四边形ADCE是平行四边形,

又∵∠ACB=90°DAB的中点,

CD= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> AB=BD=AD

平行四边形ADCE是菱形;

2)解:过点DDFCE,垂足为点F,如图所示:

DF即为菱形ADCE的高,

∵∠B=60°CD=BD

∴△BCD是等边三角形,

∴∠BDC=BCD=60°CD=BC=6

CEAB

∴∠DCE=BDC=60°

又∵CD=BC=6

RtCDF中,DF=6× <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> =3 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

17. 分析:(1)欲证明∠CEB=CBE,只要证明∠CEB=ABD,∠CBE=ABD即可.

2)先证明四边形CEDB是平行四边形,再根据BC=BD即可判定.

证明;(1)∵△ABC≌△ABD

∴∠ABC=ABD

CEBD

∴∠CEB=DBE

∴∠CEB=CBE

2))∵△ABC≌△ABD

BC=BD

∵∠CEB=CBE

CE=CB

CE=BD

CEBD

四边形CEDB是平行四边形,

BC=BD

四边形CEDB是菱形.

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

18. 分析:(1)首先根据角平分线的性质得到∠DAC=BAC,∠ABD=DBC,然后根据平行线的性质得到∠DAB+∠CBA=180°,从而得到∠BAC+∠ABD= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> (∠DAB+∠ABC= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> ×180°=90°,得到答案∠AOD=90°

2)根据平行线的性质得出ADB=DBC,∠DAC=BCA,根据角平分线定义得出∠DAC=BAC,∠ABD=DBC,求出∠BAC=ACB,∠ABD=ADB,根据等腰三角形的判定得出AB=BC=AD,根据平行四边形的判定得出四边形ABCD是平行四边形,即可得出答案.

解:(1)∵ACBD分别是∠BAD、∠ABC的平分线,

∴∠DAC=BAC,∠ABD=DBC

AEBF

∴∠DAB+∠CBA=180°

∴∠BAC+∠ABD= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> (∠DAB+∠ABC= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> ×180°=90°

∴∠AOD=90°

2)证明:∵AEBF

∴∠ADB=DBC,∠DAC=BCA

ACBD分别是∠BAD、∠ABC的平分线,

∴∠DAC=BAC,∠ABD=DBC

∴∠BAC=ACB,∠ABD=ADB

AB=BCAB=AD

AD=BC

ADBC

四边形ABCD是平行四边形,

AD=AB

四边形ABCD是菱形.

19. 分析:(1)由四边形ABCD是平行四边形,即可得AD=BCAB=CD,∠A=C,又由EF分别为边ABCD的中点,可证得AE=CF,然后由SAS,即可判定△ADE≌△CBF

2)先证明BEDF平行且相等,然后根据一组对边平行且相等的四边形是平行四边形,再连接EF,可以证明四边形AEFD是平行四边形,所以ADEF,又ADBD,所以BDEF,根据菱形的判定可以得到四边形是菱形.

解答: 1)证明:∵四边形ABCD是平行四边形,

AD=BCAB=CD,∠A=C

EF分别为边ABCD的中点,

AE= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> ABCF= <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a> CD

AE=CF

在△ADE和△CBF中,

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>

∴△ADE≌△CBFSAS);

2)若∠ADB是直角,则四边形BEDF是菱形,理由如下:

解:由(1)可得BE=DF

又∵ABCD

BEDFBE=DF

四边形BEDF是平行四边形,

连接EF,在ABCD中,EF分别为边ABCD的中点,

DFAEDF=AE

四边形AEFD是平行四边形,

EFAD

∵∠ADB是直角,

ADBD

EFBD

又∵四边形BFDE是平行四边形,

四边形BFDE是菱形.

 <a href="/tags/868/" title="判定" class="c1" target="_blank">判定</a> <a href="/tags/899/" title="菱形" class="c1" target="_blank">菱形</a>