如图所示,在△ABC中,∠ACB=90°,AD平分∠BAC交BC于D,CG⊥AB于G,交AD于F,DE⊥AB于E,那么四

∠ACB=90°,AD平分∠BAC,
所以∠CAD=∠EAD,
DE⊥AB,∠AED=90°,
∠ADC=∠ACB-∠CAD=90°-∠CAD=∠AED-∠EAD=∠ADE,
RT△ADC≌RT△ADE,[ASA],
CD=DE,AC=AE,
CG⊥AB,DE⊥AB,CG‖DE,
∠ADE=∠AFG,[同位角]
∠AFG=∠CFD,
∠ADE=∠CFD,
又∠ADE=∠ADC,所以∠ADC=∠CFD,CF=CD=DE;
∠CAD=∠EAD,
AC=AE,AF=AF,
△AFC≌△AFE,[SAS],
CF=FE=CD=DE,∠AFC=∠AFE,
∠EFD=180°-∠AFC=180°-∠AFE=∠EFD,
∠CDF=∠EAD,CD‖GE,
所以四边形CDEF是菱形.