平行四边形的性质习题(有答案)

∠C,∴∠C=80°.12.36点拨:2(AB+BC)=18,设AB=x,BC=2x,x+2x=3x=9,AB=3,BC=6,AD=BC=6cm13.150°30°140°14.4915.答案不唯一.如:BE=DF或BF=DE或∠BCE=∠DAF或AF∥EC等.16.9点拨:有ABCD,EBCF,EBNO,ONCF,AEOM,MOFD,AEFD,ABNM,MNCD.17.证明:∵四边形ABCD是平行四边形,∴AB=CD,∠B=∠D.∵AD∥BC,∴∠DEC=∠BCE.∵AF∥CE,∴∠AFB=∠BCE,∴∠DEC=∠AFB,∴△ABF≌△CDE.18.点拨:证明△ABE≌△CDF.19.9cm20.解:DE=BF.证明如下:∵O为AC的中点,∴OA=OC.又AE∥CF,∴∠EAO=∠FCO.故在△AOE与△COF中,∴△AOE≌△COF(ASA),∴AE=CF.又∵AD=CB(平行四边形的对边相等),∴AE-AD=CF-CB,即DE=BF.21.解:(1)∵ABCD,∴AB=CD,DC∥AB,∴∠ECD=∠EFA∵DE=AE,∠DEC=∠AEF∴△DEC≌△AEF∴DC=AF∴AB=AF(2)∵BC=2AB,AB=AF∴BC=BF∴△FBC为等腰三角形再由△DEC≌△AEF,得EC=EF∴∠EBC=∠EBF=∠CBF=×70°=35°22.(1)解:有4对全等三角形.分别为△AMO≌△CNO,△OCF≌△OAE,△AME≌△CNF,△ABC≌△CDA.(2)证明:如图,∵OA=OC,∠1=∠2,OE=OF.∴△OAE≌△OCF,∴∠EAO=∠FCO.在ABCD中,AB∥CD,∴∠BAO=∠DCO.∴∠EAM=∠NCF.23.(1)取AE=CF,从而可得BE=DF(或BE∥DF),证明过程略;(2)取AE=BF,可得结论四边形ABFE(或FCDE)是平行四边形,证明略.