Comparam ΔBEC cu ΔBAF
<ECB=<FAB=90
AF=CE (din ipoteza)
AB=BC
=>
ΔBEC = ΔBAF
=> BF=BE
=> ΔBFE=isoscel
=> <BFE=<BEF
in ΔDFE avem: <DEF=90-<DFE=90-<AFE
in ΔFAB avem: <AFB=<AFE+<EFB
dar <BEC=<AFB si <EFB=<FEB
=> <BEC=<FEB+<AFE
Si avem
<DEF+ <FEB+ <BEC=180
(90-<AFE)+<FEB+(<FEB+<AFE)=180
=> 90-2*FEb=180
=> <FEB=90/2=45°
=>
<EFB=<FEB=45
