Question

如图,已知正方形ABCD中,E为CB延长线上一点,F在AD边上,且BE=DF，EF与AC交于点O，求证:△OEC为等腰直角三角

详细的话+10

Answer 1

解:连接DB，∵FD∥且＝EB,

                   ∴四边形FDBE是平行四边形

           ∴∠FEC＝∠ADB＝45°

                  ∵∠AFE＝∠FEC＝45°

                    又∵∠DAC＝45°

                      ∴∠AOF=∠EOC＝180°-﹙45°＋45°﹚＝90°

                     又∵∠ACB＝45°,∠FEC＝45°

                         ∴△OEC为等腰直角三角形