如图，AC=CF,CB=CG,∠FCA=∠GCB=60∘,AE=12AG,BH=12BFAC=CF,CB=CG,\angle FCA=\angle GCB=60^{\circ},AE=\dfrac{1}{2}AG,BH=\dfrac{1}{2}BFAC=CF,CB=CG,∠FCA=∠GCB=60∘,AE=21AG,BH=21BF，求证：ΔCEH\Delta CEHΔCEH 为等边三角形
