Cho tgiac ABC với MN//BC, M,N thuộc AB,AC
Kẻ \(BH\perp AC\), ta có: \(\frac{S_{ANB}}{S_{ABC}}=\frac{\frac{1}{2}BH.AN}{\frac{1}{2}BH.AC}=\frac{AN}{AC}\left(1\right)\)
Tương tự ta cũng có: \(\frac{S_{AMC}}{S_{ABC}}=\frac{AM}{AB}\left(2\right)\)
Ta cần CM (1)=(2)
Ta có: \(S_{ANB}=S_{AMN}+S_{MNB}\left(3\right)\)
\(S_{AMC}=S_{AMN}+S_{MNC}\left(4\right)\)
Mà MN//BC nên : \(S_{MNB}=S_{MNC}\Rightarrow\left(3\right)=\left(4\right)\)
Từ đó có: \(S_{ANB}=S_{AMC}\Rightarrow\left(1\right)=\left(2\right)\Rightarrow\frac{AN}{AC}=\frac{AM}{AB}\)
