切,没过程,look
解:过点E、F分别作AB的平行线EG、FH,由平行线的传递性可得AB∥EG∥FH∥CD,
∵AB∥FH,∴∠ABF=∠BFH,
∵FH∥CD,∴∠CDF=∠DFH,
∴∠BFD=∠DFH+∠BFH=∠CDF+∠ABF;
同理可得∠BED=∠DEG+∠BEG=∠ABE+∠CDE;
∵∠ABF=2 3 ∠ABE,∠CDF=2 3 ∠CDE,
∴∠BFD=∠DFH+∠BFH=∠CDF+∠ABF=2 3 (∠ABE+∠CDE)=2 3 ∠BED,
∴∠BED:∠BFD=3:2.
3:2