在正方形ABCD的边BC的延长线上取一点E,使CE=AC,连结AE交CD于F,求∠E、∠AFC

题目

在正方形ABCD的边BC的延长线上取一点E,使CE=AC,连结AE交CD于F,求∠E、∠AFC

答案

因为 CE = AC,所以 ΔCAE 为等腰三角形,∠E = ∠CAE
∠ACB = ∠CAE + ∠E = ∠E + ∠E = 2∠E
∠E = (∠ACB) / 2 = 45° / 2 = 22.5°
∠AFC = ∠DCE + ∠E = 90° + 22.5° = 112.5°