a: sửa đề: BM cắt CN tại O
Ta có; AM+MC=AC
=>\(MC=AC-AM=AC-\frac23AC=\frac13AC\)
=>AM=2MC
=>\(S_{BMA}=2\cdot S_{BMC};S_{OMA}=2\cdot S_{OMC}\)
=>\(S_{BMA}-S_{OMA}=2\left(S_{BMC}-S_{OMC}\right)\)
=>\(S_{BOA}=2\cdot S_{BOC}\)
Ta có: AN+NB=AB
=>\(NB=AB-AN=AB-\frac13AB=\frac23AB\)
=>\(AN=\frac12NB\)
=>\(S_{CNA}=\frac12\cdot S_{CNB};S_{ONA}=\frac12\cdot S_{ONB}\)
=>\(S_{CNA}-S_{ONA}=\frac12\cdot\left(S_{CNB}-S_{ONB}\right)\)
=>\(S_{COA}=\frac12\cdot S_{COB}\)
=>\(\frac{S_{AOB}}{S_{AOC}}=2:\frac12=4\)
Ta có: \(AM=\frac23AC\)
=>\(S_{AOM}=\frac23\cdot S_{AOC}\)
=>\(\frac{S_{AOM}}{S_{AOB}}=\frac23:4=\frac{2}{12}=\frac16\)
=>\(\frac{OM}{OB}=\frac16\)
=>\(\frac{BO}{BM}=\frac67\)
b: Ta có: \(S_{AOC}=\frac14\cdot S_{AOB}\)
\(S_{AON}=\frac13\cdot S_{AOB}\left(AN=\frac13AB\right)\)
Do đó: \(\frac{S_{ANO}}{S_{AOC}}=\frac13:\frac14=\frac43\)
=>\(\frac{ON}{OC}=\frac43\)
=>\(\frac{OC}{NC}=\frac{3}{4+3}=\frac37\)
