数学作业，急高手

解：为方便计算，设三角形ABC中AB = AC =2，则BC = 2根号2，CD = 根号2，AE = EC =1在直角三角形BAE中，tan∠ABE= AE/AB = 1/2所以，∠ABE = arctan1/2根据三角形外角定理，∠ACB = ∠CAD + ∠ADC = 45度根据三角函数定理可得CD/sin∠CAD = AC/sin∠ADC2 * sin∠CAD = 根号2 * sin∠ADC =根号2 * sin(45度 - ∠CAD)化简得sin∠CAD /cos∠CAD = tan∠CAD = 1/3∠CAD = arctan1/3所以∠AFB = 180度 - ∠ABE - ∠BAF= 180度 - ∠ABE - ∠BAC - ∠CAD= 90度 - arctan1/2 - arctan1/3tan∠AFB = tan（90度 - arctan1/2 - arctan1/3）= cot（arctan1/2 + arctan1/3）= (1 – tan(arctan1/2)*tan(arctan1/3)) / (tan(arctan1/2) + tan(arctan1/3))= 1所以∠AFB = 45度 收起