Question

如图,正方形ABCD中,E为AB上一点,过点D作DF⊥DE与BC延长线交于点F,连接EF,CD交于点G,与BD交于H.

Answer 1

解

1.∵⊿BCD是等腰直角三角形.且对角线BD=√2

∴正方形ABCD的边长为 1

∵∠ADC=∠EDF=90°

∴∠ADC - ∠EDC = ∠EDF - ∠EDC

即：∠ADE = ∠CDF

又 AD = CD,∠A = ∠FCD = 90°

∴⊿ADE ≌ ⊿CDF

∴ AE = CF =BF- BC

= √2 - 1

∴ BE = AB - AE

= 1 - ( √2 - 1)

= 2 - √2

2.找不到 O 点.