设AC1和CA1相交于O点,连结EO、AE、EC1、A1E、C1E,
∵B1C1=A1B1,
A1B1=AB,
∴AB=B1C1,
∵B1E=BE.
〈ABE=〈C1B1E=90度,
∴△ABE≌△C1B1E,
∴AE=C1E,
∴△EAC1是等腰△,
∵O是AC1中点,
∴EO⊥AC1,
同理可证EO⊥A1C,
AC1∩A1C=O,
∴EO⊥平面AA1C1C,
∵EO∈平面AEC1,
∴平面AEC1⊥平面AA1C1C.