Question

如图,在四边形ABCD中,AB=BC=CD=DA,∠A=∠B=∠C=∠D=90°,点E是BC的中点，点F是CD上一点，且CF=1/4CD。求证

求证：∠AEF=90°

Answer 1

AB/CE=AB / 1/2AB = 2
BE/CF=1/2AB / 1/4AB = 2
∴ AB/CE = BE/CF
∵ ∠B=∠C=90°
∴ △ABE ∽△ECF
∠AEB=∠EFC
∠AEF = 180 -∠AEB -∠CEF = 180 - (∠EFC+∠CEF) = 180-90 = 90°

若已经学过勾股定理，用此定理证明也很方便
AE² = AB²+BE² =（5/4）AB²
EF² = CE²+CF² =（5/16）AB²
AF² = AD²+DF² = （25/16）AB
AE²+EF² =（25/16）AB² = AF²

∴ ∠AEF = 90°

Answer 2

∵E为中点
∴EC=1/2BC BE=1/2BC
∵AB=BC=CD=DA
∴EC=1/2DC BE=1/2DC
∵FC=1/4DC
∴DF=3/4DC

∵∠C=90°
∴EC²+FC²=EF²
∴EF²=5/16DC
同理可得
AE²=5/4DC AF²=25/16DC
∵5/16DC+5/4DC=25/16DC
∴AE²+EF²=AF²
∴∠AEF=90°