证明：∵△ABC为等腰△∴∠ABC=∠CAB=45度, ∠ACB=90度,AC=BC∵BF⊥AB∴∠DEB=90度∴∠EDB=45度∵BF∥AC∴∠FBC=90度∴∠DFB=45度∴∠DFB=∠EDB∴BD=BF∵D为BC中点∴CD=BD∴CD=BF∵在△ACD与△CBF中 AC=CB ∠ACD=∠CBF CD=BF∴△ACD全等于△CBF∴∠CAD=∠BCF设CF与AD交与G∵∠AGC=∠FCB+∠ADC∴∠AGC=∠CAD+∠ADC=90度