正方形ABCD中,F在BC上,E在CD上,AE=BC+CE,∠BAF=∠FAE 求证:BF=CF?答:正方形ABCD,AB=BC,在AE上找1点C',使得C'E=CE,则AC'=AE-C'E=BC+CE-C'E=BC=AB,∠BAF=∠FAE AF=AF,△AFC'≌△AFB,[SAS]FC'=FB,∠AC'F=∠B=90°.FC²=FE²-CE²=FE²-CE'²=FC'²FC=FC'=FB BF=CF.,3,过f点做ae的垂线于g则 ae=bc+...
如图,在正方形ABCD中,E是CD中点,EF⊥AE交BC于F,求证∠1=∠2答:另一解法:延长FE且与AD的延长线交于G.则因DE=EC, ∠GDE=∠FCE=90°, ∠DEG=∠CEF(对顶角),∴△GDE≌△FCE.因此有EG=EF.在△AEG和△AEF中,AE⊥FG,EG=EF,AE是FG的垂直平分线,所以∠GAE=∠FAE.又∠1、∠2分别为∠GAE、∠FAE的余角,所以有∠1=∠2....