第五章相交线与平行线专题训练(三)推理证明1.完成下面的推理过程,并在括号内填上理由:如图,已知AB∥CD,∠1=∠2,试说明EP∥FQ.证明:∵AB∥CD(________),∴∠MEB=∠MFD(___________________________),又∵∠1=∠2(________),∴∠MEB-∠1=∠MFD-∠2(等式的性质),即∠MEP=∠________(角的和差),∴EP∥______(____________________________).已知两直线平行,同位角相等已知MFQFQ同位角相等,两直线平行2.完成下列证明,并在括号内填上理由.如图,BC∥AD,∠1=∠E,求证:∠A=∠C.证明:∵∠1=∠E(已知),∴AB∥______(______________________________),∴∠A+∠ADC=180°(___________________________).∵BC∥AD(已知),∴∠ADC+∠____=180°(______________________________).∴∠A=∠C(_____________________).CE内错角相等,两直线平行两直线平行,同旁内角互补C两直线平行,同旁内角互补同角的补角相等3.完成下列证明过程,并在括号内填上依据.如图,AB∥EF,∠D=∠E,∠B+∠D=180°,求证:BC∥DE.证明:∵∠D=∠E(已知),∴CD∥____(_______________________________).又∵AB∥EF(已知),∴AB____CD(___________________________________________________________),∴∠B=∠____(____________________________).∵∠B+∠D=180°(已知),∴∠____+∠D=180°(等量代换),∴BC∥DE(_____________________________).EF内错角相等,两直线平行∥如果两条直线都与第三条直线平行,那么这两条直线也互相平行C两直线平行,内错角相等C同旁内角互补,两直线平行4.如图,AD⊥BC,FG⊥BC,∠1=∠2,求证:∠BAC=∠DEC.证明:∵AD⊥BC,FG⊥BC(已知),∴_________=_________=90°.∴AD∥FG(__________________________).∴∠1=_______(___________________________).又∵∠1=∠2(已知),∴∠2=∠____(____________).∴__________(______________________________).∴∠BAC=∠DEC(___________________________).∠ADB∠FGB同位角相等,两直线平行∠3两直线平行,同位角相等3等量代换AB∥DE内错角相等,两直线平行两直线平行,同位角相等5.完成下面的推理过程,并在括号内填上理由:如图,∠1=∠2,∠3=∠E,求证:∠A=∠EBC.证明:∵∠1=∠2(已知),∴____∥____(____________________________),∴∠E=∠____(__________________________).又∵∠E=∠3(已知),∴∠3=∠____(等量代换),∴____∥____(内错角相等,两直线平行),∴∠A=∠EBC(___________________________).DBEC内错角相等,两直线平行4两直线平行,内错角相等4ADBE两直线平行,同位角相等6.如图,∠B=∠C,∠A=∠D,求证:∠AMC=∠BND.证明:∵∠B=∠C(已知),∴AB∥______(______________________________).∴∠A=__________(两直线平行,内错角相等).又∵∠A=∠D(已知),∴∠CEA=_______(等量代换).∴______∥______(____________________________).∴∠EMB=________(两直线平行,同位角相等).又∵____________________(对顶角相等).∴∠AMC=_________(等量代换).CD内错角相等,两直线平行∠CEA∠DAEDF同位角相等,两直线平行∠BND∠EMB=∠AMC∠BND