a: Xét hình thang ABCD có EF//AB//CD
nên AE/ED=BF/FC
b: Ta có: AE/ED=BF/FC
nên ED/AE=FC/BF
\(\Leftrightarrow\dfrac{ED}{AE}+1=\dfrac{FC}{BF}+1\)
\(\Leftrightarrow\dfrac{AD}{AE}=\dfrac{BC}{BF}\)
hay AE/AD=BF/BC
c: Ta có: AE/ED=BF/FC
=>AE/ED+1=BF/FC+1
=>AD/ED=BC/CF
hay ED/DA=CF/BC