Hai tg ABM và tg ABC có chung đường cao từ A->BC nên
\(\dfrac{S_{ABM}}{S_{ABC}}=\dfrac{BM}{BC}=\dfrac{1}{3}\Rightarrow S_{ABM}=\dfrac{1}{3}S_{ABC}\)
Hai tg BCN và tg ABC có chung đường cao từ B->AC nên
\(\dfrac{S_{BCN}}{S_{ABC}}=\dfrac{CN}{AC}=\dfrac{1}{3}\Rightarrow S_{BCN}=\dfrac{1}{3}S_{ABC}\)
\(\Rightarrow S_{ABM}=S_{BCN}=\dfrac{1}{3}S_{ABC}\)
C/m tương tự ta cũng có
\(S_{APC}=S_{BCN}=\dfrac{1}{3}S_{ABC}\)
Ta có
\(S_{KIJ}=S_{ABC}-S_{ABM}-S_{CMIN}-S_{ANJK}=\)
\(=S_{ABC}-S_{ABM}-\left(S_{BCN}-S_{BIM}\right)-\left(S_{APC}-S_{APK}-S_{CJN}\right)=\)
\(=S_{ABC}-\dfrac{1}{3}S_{ABC}-\left(\dfrac{1}{3}S_{ABC}-S_{BIM}\right)-\left(\dfrac{1}{3}S_{ABC}-S_{APK}-S_{CJN}\right)=\)
\(=S_{APK}+S_{BIM}+S_{CJN}\)
