已知,如图,角ABC=角DCB,BD,CA分别是角ABC,角DCB的角平分线.求证AD//BC

连接AD∵BD、CA分别平分∠ABC和∠DCB∴∠ABO=∠OBC,∠DCO=∠OCB∵∠ABC=∠DCB∴∠ABC-∠ABO=∠DCB-∠DCO即∠OBC=∠OCB在△ABC和△DCB中∠ABC=∠DCBBC=CB∠OBC=∠OCB∴△ABC≌△DCB（ASA）∴AB=DC∴∠AC=BG ∴△ABD≌...