菱形证明专题训练

第1页共19页.............o............o............o.............o.............外.............o............o.............o.............装.............o.............o............o.............订.............o.............o.............o.............线.............o.............o..........................o.............o............o.........学校：___________姓名：____________班级：____________考号：____________.............o.............o..........................o.............内.............o.............o.............o...........装.............o.............o.............o............订.............o.............o.............o............线.............o.............o.............o.............o.............o.............o........绝密★启用前乐学教育菱形证明专题训练1.已知:如图,在四边形ABCD中,AB∥CD,E,F为对角线AC上两点,且AE=CF,DF∥BE,AC平分∠BAD.求证:四边形ABCD为菱形.【答案】∵AB∥CD,∴∠BAE=∠DCF.∵DF∥BE,∴∠BEF=∠DFE,∴∠AEB=∠CFD.又∵AE=CF,∴△AEB≌∠CFD,∴AB=CD.∵AB∥CD,∴四边形ABCD是平行四边形.∵AC平分∠BAD,∴∠BAE=∠DAF.又∠BAE=∠DCF,∴∠DAF=∠DCF,∴AD=CD,∴四边形ABCD是菱形.2.如图，矩形ABCD中，点O为AC的中点，过点O的直线分别与AB，CD交于点E，F，连接BF交AC于点M，连接DE，BO.若∠COB=60°，FO=FC.求证:(1)四边形EBFD是菱形;【答案】连接OD.∵点O为矩形ABCD的对角线AC的中点，∴B,D,O三点共线且BD=DO=CO=AO.在矩形ABCD中，AB∥DC，AB=DC，∴∠FCO=∠EAO.在△CFO和△AEO中，∴△CFO≌△AEO，∴FO=EO.又∵BO=DO，∴四边形BEFD是平行四边形.∵BO=CO，∠COB=60°，∴△COB是等边三角形.∴∠OCB=60°.第2页共19页※※请※※不※※要※※在※※装※※订※※线※※内※※答※※题※※※※※※※※※※※※※※※※※.............o.............o.............外.............o.............o.............装.............o.............o.............订.............o.............o.............线.............o.............o..............o..............o..............o.......................∴∠FCO=∠DCB-∠OCB=30°.∵FO=FC，∴∠FOC=∠FCO=30°.∴∠FOB=∠FOC+∠COB=90°.∴EF⊥BD.∴平行四边形EBFD是菱形.(2)MB∶OE=3∶2.【答案】∵BO=BC，∴点B在线段OC的垂直平分线上.∵FO=FC，∴点F在线段OC的垂直平分线上.∴BF是线段OC的垂直平分线.∴∠FMO=∠OMB=90°.∴∠OBM=30°.∴OF=BF.∵∠FOC=30°，∴FM=OF.∴BM=BF-MF=2OF-OF=OF.即FO=EO，∴BM∶OE=3∶2.3.如图,在△ABC中,∠ABC=90°,BD为AC边上的中线,过点C作CE⊥BD于点E,过点A作BD的平行线,交CE的延长线于点F,在AF的延长线上截取FG=BD,连接BG,DF.求证:四边形BGFD是菱形.【答案】∵FG∥BD,BD=FG,∴四边形BGFD是平行四边形.∵CF⊥BD,AG∥BD,∴CF⊥AG.又∵∠ABC=90°,点D是AC的中点,∴BD=DF=AC,∴平行四边形BGFD是菱形.4.如图,点O是菱形ABCD对角线的交点,DE∥AC,CE∥BD,连接OE.求证:OE=BC.【答案】∵DE∥AC,CE∥BD,∴四边形OCED是平行四边形.∵四边形ABCD是菱形,∴AC⊥BD,OB=OD,∴∠BOC=∠COD=90°,第3页共19页.............o............o............o.............o.............外.............o............o.............o.............装.............o.............o............o.............订.............o.............o.............o.............线.............o.............o..........................o.............o............o.........学校：___________姓名：____________班级：____________考号：____________.............o.............o..........................o.............内.............o.............o.............o...........装.............o.............o.............o............订.............o.............o.............o............线.............o.............o.............o.............o.............o.............o........∴四边形OCED是矩形,∴∠ODE=90°,∵OB=OD,∠BOC=∠ODE=90°,∴BC=,OE=,∵DE=OC.∴OE=BC.5.[2015·兰州中考,25](9分)如图,四边形ABCD中,AB∥CD,AB≠CD,BD=AC.(1)求证:AD=BC;【答案】作BM∥AC,BM交DC的延长线于点M,则∠ACD=∠BMD.1分∵AB∥CD,BM∥AC,∴四边形ABMC为平行四边形.2分∴AC=BM.∵BD=AC,∴BM=BD.∴∠BDM=∠BMD.∴∠BDC=∠ACD.在△BDC和△ACD中,∴△BDC≌△ACD.4分∴BC=AD.5分(2)若E,F,G,H分别是AB,CD,AC,BD的中点.求证:线段EF与线段GH互相垂直平分.【答案】连接EG,GF,FH,HE.6分∵E,H为AB,BD的中点,∴EH=AD.同理FG=AD,EG=BC,FH=BC.∵BC=AD,∴EG=FG=FH=EH.8分∴四边形EGFH为菱形,∴EF与GH互相垂直平分.9分第4页共19页※※请※※不※※要※※在※※装※※订※※线※※内※※答※※题※※※※※※※※※※※※※※※※※.............o.............o.............外.............o.............o.............装.............o.............o.............订.............o.............o.............线.............o.............o..............o..............o..............o.......................6.[2015·长春中考,18](7分)如图,CE是△ABC外角∠ACD的平分线,AF∥CD交CE于点F,FG∥AC交CD于点G,求证:四边形ACGF是菱形.【答案】因为AF∥CD,FG∥AC,所以四边形ACGF是平行四边形①,又因为∠ACE=∠ECG,∠ECG=∠AFC,所以∠ACE=∠AFC,所以AC=AF②,由①②得四边形ACGF是菱形.7.[2010·上海中考，23]已知梯形ABCD中，AD∥BC，AB＝AD(如图所示)，∠BAD的平分线AE交BC于点E，连结DE.(1)在图中，用尺规作∠BAD的平分线AE(保留作图痕迹，不写作法)，并证明四边形ABED是菱形；【答案】∵∠BAE＝∠DAE，∠DAE＝∠BEA，∴∠BAE＝∠BEA，AB＝BE＝AD，AD∥BE，∴四边形ABED的平行四边形，又AB＝AD，∴四边形ABED为菱形(2)∠ABC＝60°，EC＝2BE，求证：ED⊥DC.【答案】过D作DF∥AE，则DF＝CF＝1，∴∠C＝30°，而∠DEC＝60°，∴∠EDC＝90°，∴ED⊥DC.第5页共19页.............o............o............o.............o.............外.............o............o.............o.............装.............o.............o............o.............订.............o.............o.............o.............线.............o.............o..........................o.............o............o.........学校：___________姓名：____________班级：____________考号：____________.............o.............o..........................o.............内.............o.............o.............o...........装.............o.............o.............o............订.............o.............o.............o............线.............o.............o.............o.............o.............o.............o........8.[2010·沈阳中考，19]如图，菱形ABCD的对角线AC与BD相交于O，点E，F分别为边AB，AD的中点，连接EF，OE，OF，求证：四边形AEOF是菱形.【答案】∵点E，F分别为AB，AD的中点∴AE＝AB，AF＝AD(2分)又∵四边形ABCD是菱形∴AB＝AD∴AE＝AF(4分)又∵菱形ABCD的对角线AC与BD相交于点O∴O为BD的中点∴OE，OF是△ABD的中位线(6分)∴OE∥AD，OF∥AB∴四边形AEOF是平行四边形(8分)第6页共19页※※请※※不※※