a, DF//BC
=> \(\frac{MD}{MF}\) = \(\frac{CE}{CF}\) = \(\frac{CE}{BD}\) (CF =BD)
= \(\frac{AC}{AB}\)
=> \(\frac{MD}{MF}\) = \(\frac{AC}{AB}\) (đpcm)
b, \(\frac{DE}{BC}\) = \(\frac{AD}{AB}\) = \(\frac{AD}{AD + BD}\)
hay \(\frac{3}{8}\) = \(\frac{AD}{AD+ 5}\)
=> 3.(AD + 5)= 8 . AD
=> 15 = 5 AD
=> AD = 3 (cm)
=> AB = AD + DB = 3 + 5 = 8 (cm) = BC
Vậy tam giác ABC cân tại B