Question

如图,正方形ABCD中,E是AD的中点,BD与CE交于点F 如图,正方形ABCD中,E是AD的中点,BD与CE交于点F,

Answer 1

设AF与BE相交于M,

DA=DC, ∠ADF=∠CDF=45°, FD=FD ==> △DAF≌△DCF ==> ∠DAF=∠DCF

AE=ED, ∠BAE=∠CDE=90°, AB=DC ==> △ABE≌△DCE ==> ∠BEA=∠CED

故∠DAF+∠BEA = ∠DCF+∠CED = 180°- ∠CDE = 90°

即∠EAM+∠MEA = 90°, 所以 ∠EMA = 180°- 90°= 90°,

即AF⊥BE.

佩服 shuxpp,但是他解答最后一步有问题,所以补充回答.