Nối MC. Chọn BMC là tam giác trung gian.
Ta có : BN = 2NC => BN = 2/3BC
=> \(\frac{S_{BMN}}{S_{BMC}}=\frac{BN}{NC}=\frac{2}{3}\)
Lại có AM = 2BM => BM = 1/3AB
\(\Rightarrow\frac{S_{BMC}}{S_{ABC}}=\frac{BM}{AB}=\frac{1}{3}\Rightarrow S_{BMC}=\frac{1}{3}S_{ABC}\) mà \(S_{BMN}=\frac{2}{3}S_{BMC}\Rightarrow S_{BMN}=\frac{2}{3}.\frac{1}{3}S_{ABC}=\frac{2}{9}S_{ABC}\)
Tương tự, ta cũng chứng minh được \(S_{AMP}=\frac{2}{9}S_{ABC};S_{PNC}=\frac{2}{9}S_{ABC}\)
\(\Rightarrow S_{AMP}+S_{BMN}+S_{PNC}=\frac{2}{9}.3S_{ABC}=\frac{2}{3}S_{ABC}\)
\(\Rightarrow S_{MNP}=S_{ABC}-\left(S_{AMP}+S_{BMN}+S_{PCN}\right)=S_{ABC}-\frac{2}{3}S_{ABC}=\frac{1}{3}S_{ABC}\)