Question

四边形ABCD是正方形,点E是边BC的中点,角AEF=90度,且EF交正方形外角的平分线CF于点F,求证AE=EF

Answer 1

取AB中点P,则AP=CE=AB/2=BC/2,PB=PE

RT△PBE中∠EPB=∠PEB=45°

所以有∠APE=180°-45°=135°

∠FCE=∠DCB°+45=135°=∠APE

∠EAB=90°-∠AEB=180°-90°-∠AEB=∠FEC

△AEP≌△EFC

所以AE=EF