的平方+的平方=0
得：ab = 1 a = b
解得：a = b = 1
所以△ABC是等腰三角形,AC = BC=1
在AB上取一点E,使得AE = AC,连接ED.
由AC+CD = AB知 CD =BE
在△CAD和△EAD中
AC = AE,
∠CAD = ∠EAD
AD =AD
所以△CAD≌△EAD ,∠ACD = ∠AED,所以ED=CD = BE.
所以∠B = ∠EDB,所以∠AED= 2∠B
又∠ACD =180° - 2 ∠B
所以180° - 2 ∠B = 2 ∠B 解得：∠B= 45°
所以∠CAB= ∠B = 45°,∠ACB =90°
所以S△ABC = AC*BC/2 = 1/2