解：在△ADB与△BCA中∵∠C=∠DAC=BDAB=BA∴△ADB≌△BCA∴AD=BC∵∠DOA与∠COB为对顶角∴∠DOA=∠COB在△DOA与△COB中∵∠COB=∠DOAAD=BC∠C=∠D∴△DOA≌△COB∴OC=OD
