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1、平行线理由填写练习1.如图1,1=2,( 已知 )_( )3=4,( 已知 )_( )5=B,( 已知 )_( )D+DAB=180,(已知 )_( )2、如右图,12 , 2 ,同位角相等,两直线平行34180 , ACFG, 3、如图2,1+2=180(已知)3+2=180( )1=_ABCD( )4、如图,BE平分ABC(已知1=3( )又1=2(已知) _=2( )_( )AED=_( )5.如图9,ABCD (已知 ) A+_=180( )BCAD,(已知 )A+_=180( )B=_.( )6、已知,如图4,1=ABC=ADC,3=5,2=4,ABC+BCD=180(1) 1=AB
2、C(已知)AD ( )(2)3=5(已知)AB ( )(3)2=4(已知) ( )(4)1=ADC(已知) ( )(5)ABC+BCD=180(已知) ( ) 7、如图5,(1)A= (已知)ACED( )(2)2= (已知)ACED( )(3)A+ =180(已知)ABFD( )(4)AB (已知)2+AED=180( )(5)AC (已知)C=1( )8、如图6:(1) EFAB,(已知) 1= ( );(2) 3= (已知) ABEF ( );(3) A= (已知) ACDF ( );(4) 2+ =1800(已知) DEBC ( );(5) ACDF(已知) 2= ( );(6) EF
3、AB(已知) FCA+ =1800( )9、如图,已知AGDACB,12。求证:CDEF。(填空并在背面旳括号中填理由)证明:AGDACB()DG( )3( )12()3( )( )10、如图,A60,DFAB于F,DGAC交AB于G,DEAB交AC于E。求GDF旳度数。解:DFAB()DFA90()DEAB()1()EDF180DFA1809090()DGAC()2()GDF()11、如图,已知BAF50,ACE140,CDCE,能判断DCAB吗?为何?解:能判断DCAB。CDCE(已知)DCE( )ACD360DCEACE36090140130CAB180BAF18050130(邻补角定义
4、)ACD( )()ABCDGHEF12、已知:如图,ABCD,EF分别交于AB、CD于E、F,EG平分AEF,FH平分EFD。求证: EGFH证明: ABCD(已知) AEF=EFD (_ _) EG平分AEF,FH平分EFD(_ _ _), _ _=AEF, _ _=EFD(角平分线定义) _ _=_ ( ) EGFH(_ _)13、已知,如图7,B =C,1 = 3,求证:A =D证明: B =C (已知) ABCD A = 又 1 = ( ) 2 = 3 ( ) 1 = 2 ( ) ( ) =D( ) A =D ( )14、如图,已知AB/DE,B=E,求证:BC/EFCO证明: AB/
5、DE ( )E B= ( )又B=E( ) = ( ) / ( )15、已知,如图,1=120,2=120,求证:AB/CD。证明:1=120,2=120( ) 1=2( )又 = ( ) 1=3( )2AB/CD( )16、已知,如图,AB/CD,BC/AD,3=4。求证:1=2A证明:AB/CD( )4321DCB = ( )又 BC/AD( ) = ( )又3=4( )1=2( )17、如上图左,已知,ADCABC,BE、DF分别平分ABC、ADC,且1=2,求证:A=C.证明:BE、DF分别平分ABC、ADC(已知) 1ABC,3ADC( )ABCADC(已知)ABCADC( )13( )12(已知)23( )( )( )( )A 180 ,C 180( )AC( )18、如图,完毕下列推理过程CBAFEDO已知:DEAO于E, BOAO,CFB=EDO证明:CFDO 证明:DEAO, BOAO(已知)DEA=BOA=900 ( )DEBO ( )EDO=DOF ( )又CFB=EDO( )DOF=CFB( )CFDO( )