skyflyfeng
幼苗
共回答了26个问题采纳率:92.3% 举报
∵EF∥AB∴∠2=∠ABC∵BF∥AC∴∠1=∠C又∵∠2=∠ABC∴⊿ABC∽⊿FEB∴∠F=∠A=30°,AB∶EF=BC∶BE=2∶1又∵EF=4∴AB=2EF=8. 不用相似也可以:如图,延长FE交AC于D.
∵AF∥BC,AB∥EF∴有平行四边形ABDF∴AB=DF=DE+EF,∠B=∠F=30°∵AE=EC,∠AEF=∠CED,∠C=∠CAF(AF∥BC)∴⊿AEF≌⊿CED∴DE=EF又AB=DF=DE+EF∴AB=2EF=8.
1年前
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