1) Ta c/m t/c sau (để áp dụng vào câu 1)
△DEF có G thuộc DE, H thuộc DF. C/m: \(\dfrac{S_{DGH}}{S_{DEF}}=\dfrac{DH}{DF}.\dfrac{DG}{DC}\)
-Ta có: \(\dfrac{S_{DGH}}{S_{DGF}}=\dfrac{DH}{DF};\dfrac{S_{DGF}}{S_{DEF}}=\dfrac{DG}{DC}\).
-Nhân vế theo vế ta có đpcm.
-Quay lại bài toán:
\(BM=2CM\Rightarrow\dfrac{BM}{BC}=\dfrac{1}{3};\dfrac{CM}{BC}=\dfrac{2}{3}\)
\(CN=3NA\Rightarrow\dfrac{CN}{CA}=\dfrac{3}{4};\dfrac{NA}{CA}=\dfrac{1}{4}\)
\(AP=BP\Rightarrow\dfrac{AP}{AB}=\dfrac{BP}{AB}=\dfrac{1}{2}\)
\(\dfrac{S_{PBM}}{S}+\dfrac{S_{CMN}}{S}+\dfrac{S_{APN}}{S}=\dfrac{BP}{AB}.\dfrac{BM}{BC}+\dfrac{CN}{CA}.\dfrac{MC}{BC}+\dfrac{AP}{AB}.\dfrac{AN}{AC}=\dfrac{1}{2}.\dfrac{2}{3}+\dfrac{3}{4}.\dfrac{1}{3}+\dfrac{1}{2}.\dfrac{1}{4}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{S-S_{MNP}}{S}=\dfrac{3}{4}\Rightarrow1-\dfrac{S_{MNP}}{S}=\dfrac{3}{4}\Rightarrow\dfrac{S_{MNP}}{S}=\dfrac{1}{4}\Rightarrow S_{MNP}=\dfrac{S}{4}\)
