a: Ta có: AM+MB=AB
=>\(MB=AB-AM=AB-\dfrac{1}{3}AB=\dfrac{2}{3}AB\)
\(\dfrac{AM}{MB}=\dfrac{\dfrac{1}{3}AB}{\dfrac{2}{3}AB}=\dfrac{1}{3}:\dfrac{2}{3}=\dfrac{1}{2}\)
=>\(\dfrac{S_{AMC}}{S_{BMC}}=\dfrac{1}{2}\)
b: \(\dfrac{MB}{AB}=\dfrac{2}{3}\)
=>\(\dfrac{S_{BMC}}{S_{ABC}}=\dfrac{2}{3}\)
c: Ta có: \(\dfrac{S_{BMC}}{S_{ABC}}=\dfrac{2}{3}\)
=>\(S_{MBC}=\dfrac{2}{3}\times48=32\left(cm^2\right)\)
N là trung điểm của BC
=>\(S_{MNC}=\dfrac{1}{2}\times S_{MBC}=\dfrac{1}{2}\times32=16\left(cm^2\right)\)