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如图,△ABC是等边三角形,AE=BD,AD与CE交于点F。求∠CFD的度数。
参考答案:\(\because \Delta ABC\)是等边三角形\(\therefore AB = CA\),\(\angle B = \angle CAB = 60^\circ \)\(\therefore \Delta ABD \cong \Delta CAE\)\(\therefore \angle BAD = \angle ACE\)\(\because \angle BAD + \angle DAC = \angle BAC = 60^\circ \)\(\therefore \angle ACE + \angle DAC = 60^\circ \)\(\because \angle DFC = \angle ACE + \angle DAC\)\(\therefore \angle DFC = 60^\circ \)。