题目内容：

已知:如图,在三角形ABC中,点E在边AC上,角AEB=角ABE+角EBC
（2）若角BAE的平分线AF交BE于点F,FD//BC,交AC于点D,且AB=15,DC=13,求AC的长

优质解答

∵∠AEB = ∠EBC + ∠C（外角=与之不相邻的两个内角和）∵已知∠AEB = ∠ABE + ∠EBC∴∠C = ∠ABE∵FD//BC∴∠C = ∠ADF∴∠ADF = ∠ABE∵AF是∠BAE的角平分线∴∠DAF = ∠BAF∵AF = AF∴△ADF≌△ABF∴AD = AB = 1...