答:AE∥BC
证明:∵∠AFD=75°,∠ADF=90°
∴∠DAF=180° - ∠AFD - ∠ADF
=180° - 75° - 90°
=15°
∵∠DAE=45°
∴∠EAF=∠DAE - ∠DAF
=45° - 15°
=30°
又∵∠C=30°
∴∠C=∠EAF
∴AE∥BC(内错角相等,两直线平行)
答:AE∥BC
证明:∵∠AFD=75°,∠ADF=90°
∴∠DAF=180° - ∠AFD - ∠ADF
=180° - 75° - 90°
=15°
∵∠DAE=45°
∴∠EAF=∠DAE - ∠DAF
=45° - 15°
=30°
又∵∠C=30°
∴∠C=∠EAF
∴AE∥BC(内错角相等,两直线平行)