结论：AB=AD+CD证明：在AB上截取AF=AC；AE=AD;连接DE;DF∵AC=BC‘∴∠BAC=∠CAB=180°-∠C∵∠C=100°；∴∴∠BAC=∠ABC=40°；∵AD平分∠CAB;∴∠DAB=20°；∵AD=AD;∴∠ADE=∠AED=`1/2(180-20)=80°；∵∠AED=∠DE...