Question

如图,△ABC中,∠CAB=90°,CB的垂直平分线交BC于点E,交CA的延长线于点D

Answer 1

∵∠BAC=90°,∴∠B+∠C=90°,

∵FE⊥BC,BE=CE,

∴∠F+∠C=90°,AE=BE,

∴∠B=∠F=∠DAE,又∠CEF为公共角,

∴ΔEAD∽ΔEFA,

∴DE/AE=AE/EF,

∴AE^2=EF*ED.