证明：将AC与BD的交点设为O∵AB＝AD∴∠ABD＝∠ADB∵∠CBD＝∠ABC-∠ABD,∠CDB＝∠ADC-∠ADB,∠ABC＝∠ADC∴∠CBD＝∠CDB∴BC＝DC∴△ABC≌△ADC （SAS）∴∠BAC＝∠DAC∵AO＝AO∴△ABO≌△ADO （SAS）∴BO＝DO,∠AO...