因为AB = AC,所以∠B = ∠C
所以∠CAD = ∠B+∠C = 2∠B
因为AD = AE,所以∠D = ∠AED
所以∠D = (180°-∠CAD)/2 = (180°-2∠B)/2 = 90° - ∠B
△DBF中,∠B + ∠D = ∠B + 90° - ∠B = 90°,所以∠DFB = 180° - 90° = 90°
所以DF⊥BC
证明:∵AB=AC
∴∠B=∠C
∴∠CAF=∠B+∠C=2∠B
∵AD=AE
∴∠D=∠AED
∴∠BAC=∠D+∠AED=2∠D
∵∠BAC+∠CAD=180°
∴2∠D+2∠B=180°
∴∠D+∠B=90°
∴∠BFD=90°
∴DF⊥BC