Hình vẽ bạn thay điểm P thành điểm K nhé.
Ta có:
\(\frac{S_{BOC}}{S_{ABC}}=\frac{\frac{1}{2}BC.OM}{\frac{1}{2}BC.AM}\)
\(\Rightarrow\frac{S_{BOC}}{S_{ABC}}=\frac{OM}{AM}.\)
Lại có:
\(\frac{S_{AOC}}{S_{ABC}}=\frac{\frac{1}{2}ON.CM}{\frac{1}{2}BN.CM}\)
\(\Rightarrow\frac{S_{AOC}}{S_{ABC}}=\frac{\frac{1}{2}ON}{\frac{1}{2}BN}\)
\(\Rightarrow\frac{S_{AOC}}{S_{ABC}}=\frac{ON}{BN}.\)
Có:
\(\frac{S_{AOB}}{S_{ABC}}=\frac{\frac{1}{2}OK.AB}{\frac{1}{2}CK.AB}\)
\(\Rightarrow\frac{S_{AOB}=\frac{1}{2}OK}{S_{ABC}=\frac{1}{2}CK}\)
\(\Rightarrow\frac{S_{AOB}}{S_{ABC}}=\frac{OK}{CK}.\)
\(\Rightarrow\frac{OM}{AM}+\frac{ON}{BN}+\frac{OK}{CK}=\frac{S_{BOC}}{S_{ABC}}+\frac{S_{AOC}}{S_{ABC}}+\frac{S_{AOB}}{S_{ABC}}\)
\(\Rightarrow\frac{OM}{AM}+\frac{ON}{BN}+\frac{OK}{CK}=\frac{S_{BOC}+S_{AOC}+S_{AOB}}{S_{ABC}}\)
\(\Rightarrow\frac{OM}{AM}+\frac{ON}{BN}+\frac{OK}{CK}=\frac{S_{ABC}}{S_{ABC}}\)
\(\Rightarrow\frac{OM}{AM}+\frac{ON}{BN}+\frac{OK}{CK}=1\left(đpcm\right).\)
Chúc bạn học tốt!
