Question

[Hiếu Minh]

Cho tam giác ABC qua một điểm O tùy ý nằm bên trong tam giác dựng các đường thẳng AO,BO,CO cắt BC,AC,AB tương ứng tại M,N,K

CMR: \(\frac{OM}{AM}+\frac{ON}{BN}+\frac{IK}{CK}=1\)

Answer 1

[Vũ Minh Tuấn]

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!