由已知:AC=AB,所以∠CAB=∠CBA
设∠=CAB=∠CBA=2x,因为AE是角平分线,所以∠CAE=∠EAB=x,
∠DEA为△AEB的外角,因为AD垂直BC,∠ADB=90°,所以∠CBA+∠EAB=∠AED
所以∠ADB+∠EAD+∠AED=180°
x+18度+3x=180°
x=41°
∠ACD=180-41-90=49°
由已知:AC=AB,所以∠CAB=∠CBA
设∠=CAB=∠CBA=2x,因为AE是角平分线,所以∠CAE=∠EAB=x,
∠DEA为△AEB的外角,因为AD垂直BC,∠ADB=90°,所以∠CBA+∠EAB=∠AED
所以∠ADB+∠EAD+∠AED=180°
x+18度+3x=180°
x=41°
∠ACD=180-41-90=49°