一楼跳过了最关键的部分,即怎么证∠BCA=∠DBC .二楼思路正确,但还是跳过了一步,而且字母标乱了.  如图,∵AD//BC,∴∠CAD=∠ACB,∠BDA=∠CBD又∵∠AOD=∠COB,∴△AOD相似于△COB,（注意字母顺序）∴OA/OC=OD/OB∴OA/(OA+OC)=OD/(OD+OB),即OA/AC=OD/BD又∵AC=BD∴OA=OD∴∠CAD=∠BDA=∠DBC=∠ACB∴在△ABC和△DCB中,有AC=BD,∠DBC=∠ACB,BC共用,∴△ABC和△DCB全等,（两边夹一角）∴AB=DC