a: Ta có: \(AN=\frac13\times AC\)
=>\(S_{BNA}=\frac13\times S_{ABC}\) (1)
Ta có: \(AM=\frac13\times AB\)
=>\(S_{AMC}=\frac13\times S_{ABC}\) (2)
Từ (1),(2) suy ra \(S_{BNA}=S_{AMC}=\frac13\times S_{ABC}\)
b: Ta có: \(AM=\frac13\times AB\)
=>\(S_{AMN}=\frac13\times S_{ANB}=\frac13\times\frac13\times S_{ABC}=\frac19\times S_{ABC}\)
Ta có: \(S_{AMN}+S_{BMNC}=S_{ABC}\)
=>\(S_{BMNC}=S_{ABC}-\frac19\times S_{ABC}=\frac89\times S_{ABC}=\frac89\times36=32\left(\operatorname{cm}^2\right)\)