(3)如图2AB=AC∠B=∠BAC=∠FEBADE≌△ADC ∠ACD=∠AED ∠FCA=90-∠ACD ∠FEB=90-∠AED∠FCA=∠FEB=∠B=∠BAC∠ACD=∠B+∠BAC=2∠FCA=90-∠FCA∠FCA=30=∠FEBFG==1÷√3