Bài 4:
2: \(CN=\dfrac{1}{2}BN\)
=>\(CN=\dfrac{1}{3}BC\)
=>\(S_{ANC}=\dfrac{1}{3}\cdot S_{ABC}\)
Vì AM=1/3AC
nên \(S_{AMN}=\dfrac{1}{3}\cdot S_{ANC}\)
=>\(S_{CMN}=\dfrac{2}{3}\cdot S_{ANC}=\dfrac{2}{3}\cdot\dfrac{1}{3}\cdot S_{ABC}=\dfrac{2}{9}\cdot S_{ABC}\)