1）过D作DE‖BC,交AB于点E.∵BE‖CD,DE‖BC,∴四边形CDEB为平行四边形,∴BE＝CD＝b,且∠CDE＝∠B＝1/2∠ADC,∴DE平分∠ADC,∴∠ADE＝∠CDE＝∠B,∵AB‖CD,∴∠AED＝∠CDE＝∠B,∴∠ADE＝∠AED,∴AE＝AD＝a,∴AB＝AE...