这道数学题不会，求解，谢谢了
解:(1)设AE=DE=5x,则AF=DF=4x,BE=6-5x,CF=6-4x.
∵∠EDF=∠A=60°.
∴∠CDF+∠BDE=∠BED+∠BDE=120°,则∠CDF=∠BED;
又∠C=∠B=60°.故⊿DCF∽⊿EBD,CD/BE=DF/ED.
CD/(6-5x)=4x/5x=4/5,CD=24/5-4x.
作DM垂直CA于M,∠CDM=30°,则CM=CD/2=12/5-2x,DM=√3CM=12√3/5-2√3x.
FM=CF-CM=18/5-2x.因FM²+DM²=DF²,即(18/5-2x)²+(12√3/5-2√3x)²=(4x)².
解得:x=7/10.则CD=24/5-4*(7/10)=2,BD=BC-CD=4.
(2)作AH垂直BC于H,则BH=3,AH=3√3;又ED垂直BC.
∴⊿BDE∽⊿BHA,DE/HA=BE/BA,AE/(3√3)=(6-AE)/6,AE=12√3-18=DE;
∠BED=30°,可得:BD=12-6√3,BE=24-12√3;CD=6√3-6.
∵∠CDF=∠BED=30°,则DF垂直AC.
∴CF=CD/2=3√3-3,DF=√3CF=9-3√3.
又∠EDF=∠EAF=60°,作EN垂直DF于N,则DN=DE/2=6√3-9,NF=DF-DN=18-9√3;
EN=√3DN=18-9√3,故EF=√2NF=18√2-9√6.得BE/EF=(24-12√3)/(18√2-9√6)=(2√2)/3.
(3)∵∠B=∠EDF=60°;若⊿BED与⊿DEF相似.
∴∠BED=∠DFE或∠BED=∠DEF.
①当∠BED=∠DFE时,又⊿EBD∽⊿DCF,∠BED=∠CDF.
∴∠DFE=∠CDF,则EF平行BC;又EF垂直平分AD,故E为AB中点,BE=BA/2=3;
②当∠BED=∠DEF时,又∠DEF=∠AEF.
∴∠BED=∠DEF=∠AEF=60°,则EF平分BC,同样可得E为AB中点,BE=BA/2=3.
