设∠B=x，∵∠DCB=2∠B，∴∠DCB=2x，∵∠C的平分线交AB于D，∴∠ACD=∠DCB=2x，∵∠ADC是△BCD的外角，∴∠ADC=∠B+∠DCB=3x，在△ACD中，∵∠A+∠ACD+∠ADC=180°，∴90°+2x+3x=180°，解得x=18°，∴∠ADC=3x=3...