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Câu 1:
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
vecto AN+vecto BP+vecto CM
=vecto AB+vecto BN+vecto BC+vecto CP+vecto CA+vecto AM
=vecto AB+1/3vecto BC+vecto BC+1/3vecto CA+vecto CA+1/3vecto AB
=4/3 vecto AB+4/3vecto BC+4/3vecto CA
=vecto 0
Ta có :
\(\overrightarrow{BP}+\overrightarrow{AN}+\overrightarrow{CM}=\overrightarrow{BC}+\overrightarrow{CP}+\overrightarrow{AB}+\overrightarrow{BN}+\overrightarrow{CA}+\overrightarrow{AM}=\overrightarrow{CP}+\overrightarrow{BN}+\overrightarrow{AM}\)\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{BC}+\dfrac{1}{3}\overrightarrow{CA}\)
\(=\dfrac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}\right)\)
\(=\dfrac{1}{3}\overrightarrow{0}\)
\(=\overrightarrow{0}\)
\(\RightarrowĐPCM\)
Vì MB = 2MC (M thuộc đoạn BC)
\(\Rightarrow\overrightarrow{MB}=-2\overrightarrow{MC}\\ \Leftrightarrow\overrightarrow{MA}+\overrightarrow{AB}=-2\overrightarrow{MA}-2\overrightarrow{AC}\\ \Leftrightarrow3\overrightarrow{AM}=\overrightarrow{AB}+2\overrightarrow{AC}\\ \Leftrightarrow\overrightarrow{AM}=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
Vì \(\frac{MA}{MB}=\frac25\)
nên \(S_{CMA}=\frac25\times S_{CMB};S_{IMA}=\frac25\times S_{IMB}\)
=>\(S_{CMA}-S_{IMA}=\frac25\times\left(S_{CMB}-S_{IMB}\right)\)
=>\(S_{CIA}=\frac25\times S_{CIB}\)
Vì \(\frac{BN}{NC}=\frac13\)
nên \(S_{ANB}=\frac13\times S_{ANC};S_{INB}=\frac13\times S_{INC}\)
=>\(S_{ANB}-S_{INB}=\frac13\times\left(S_{ANC}-S_{INC}\right)\)
=>\(S_{AIB}=\frac13\times S_{AIC}\)
=>\(S_{AIB}=\frac13\times\frac25\times S_{CIB}=\frac{2}{15}\times S_{CIB}\)
Vì \(\frac{BN}{NC}=\frac13\)
nên NC=3BN
Ta có: NC+BN=BC
=>BC=3BN+BN=4BN
=>\(BN=\frac14\cdot BC\)
=>\(S_{INB}=\frac14\cdot S_{IBC}\)
=>\(\frac{S_{AIB}}{S_{IBN}}=\frac{2}{15}:\frac14=\frac{8}{15}\)
=>\(\frac{AI}{IN}=\frac{8}{15}\)
=>\(\frac{AI}{AN}=\frac{8}{23}\)
Vì \(\frac{AM}{MB}=\frac25\)
nên \(AM=\frac25MB\)
Ta có: AM+MB=AB
=>\(AB=\frac25MB+MB=\frac75MB\)
=>\(MB=\frac57BA\)
=>\(S_{BMI}=\frac57\cdot S_{BIA}=\frac57\cdot\frac{2}{15}\cdot S_{BIC}=\frac{10}{105}\cdot S_{BIC}=\frac{2}{21}\cdot S_{BIC}\)
=>\(\frac{MI}{IC}=\frac{2}{21}\)
=>\(\frac{CI}{IM}=\frac{21}{2}\)