NA+NB=AB
=>AB=5NB+NB=6NB
=>\(AN=\frac56AB;BN=\frac16AB\)
AP+PC=AC
=>AC=10PC+PC=11PC
=>\(CP=\frac{1}{11}AC;AP=\frac{10}{11}AC\)
M đối xứng B qua C
=>C là trung điểm của BM
=>MB=2CB
\(\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{BN}=2\cdot\overrightarrow{CB}+\frac16\cdot\overrightarrow{BA}\)
\(=2\cdot\overrightarrow{CA}+2\cdot\overrightarrow{AB}-\frac16\cdot\overrightarrow{AB}=\frac{11}{6}\cdot\overrightarrow{AB}-2\cdot\overrightarrow{AC}\)
\(=\frac{11}{6}\left(\overrightarrow{AB}-\frac{12}{11}\cdot\overrightarrow{AC}\right)\) (1)
\(\overrightarrow{MP}=\overrightarrow{MC}+\overrightarrow{CP}=\overrightarrow{CB}+\frac{1}{11}\cdot\overrightarrow{CA}\)
\(=\overrightarrow{CA}+\overrightarrow{AB}+\frac{1}{11}\cdot\overrightarrow{CA}=\overrightarrow{AB}-\frac{12}{11}\cdot\overrightarrow{AC}\) (2)
Từ (1),(2) suy ra \(\overrightarrow{MN}=\frac{11}{6}\cdot\overrightarrow{MP}\)
=>M,N,P thẳng hàng
