a) Vì BE//AC
=> \(\widehat{A1}=\widehat{E2} \) (2 góc slt)
Xét ΔADMvàΔEDB có:
\(\widehat{A1}=\widehat{E2} \) (cmtrn)
\(\widehat{D1}=\widehat{D2} \) (2 góc đối đỉnh)
=> ΔADM∼ΔEDB (g.g)
b) Theo câu a) => \(\frac{BE}{AM}=\frac{ED}{AD}=\frac{DM}{DB}\)
( mà DM=DB=\(\frac{1}{2}MB\))
=> \(\frac{BE}{AM}=\frac{ED}{AD}=\frac{DM}{DB}=1\)