延长正方形ABCD的BC边至点E,使CE=AC,连结AE,交CD于F,那么∠AFC的度数为（ ）,若BC=4cm,则S△ACE=?
问题描述：
延长正方形ABCD的BC边至点E,使CE=AC,连结AE,交CD于F,那么∠AFC的度数为（ ）,若BC=4cm,则S△ACE=?
答
因为∠E = CAE
外角∠ACB = ∠E + CAE = 45°
所以2∠E = 45°
∠E = 22.5°
∠AFC=∠FCE + ∠E = 90°+ 22.5° = 112.5°
CE = AC = 4√2
S△ACE= = 1/2 CE * AB = 1/2 * 4√2 * 4 = 8 √2 (cm²)