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BM=MC
=>\(S_{AMB}=S_{AMC};S_{OMB}=S_{OMC}\)
=>\(S_{AMB}-S_{OMB}=S_{AMC}-S_{OMC}\)
=>\(S_{AOB}=S_{AOC}\)
Ta có: \(AN=\frac13\times NC\)
=>\(S_{BNA}=\frac13\times S_{BNC};S_{ONA}=\frac13\times S_{ONC}\)
=>\(S_{BNA}-S_{ONA}=\frac13\times\left(S_{BNC}-S_{ONC}\right)\)
=>\(S_{BOA}=\frac13\times S_{BOC}\)
=>\(S_{COA}=\frac13\times S_{COB}\)
Ta có; P nằm giữa A và B
=>\(\frac{S_{CPA}}{S_{CPB}}=\frac{PA}{PB};\frac{S_{OPA}}{S_{OPB}}=\frac{PA}{PB}\)
=>\(\frac{PA}{PB}=\frac{S_{CPA}-S_{OPA}}{S_{CPB}-S_{OPB}}=\frac{S_{COA}}{S_{COB}}=\frac13\)
a: Ta có: \(AM=\frac12MB\)
=>\(S_{CMA}=\frac12\times S_{CMB};S_{PMA}=\frac12\times S_{PMB}\)
=>\(S_{CMA}-S_{PMA}=\frac12\times\left(S_{CMB}-S_{PMB}\right)\)
=>\(S_{CPA}=\frac12\times S_{CPB}\)
Ta có: \(AN=\frac13NC\)
=>\(S_{BNA}=\frac13\times S_{BNC};S_{PNA}=\frac12\times S_{PNC}\)
=>\(S_{BNA}-S_{PNA}=\frac13\times\left(S_{BNC}-S_{PNC}\right)\)
=>\(S_{BPA}=\frac13\times S_{BPC}\)
TA có: \(S_{APB}+S_{BPC}+S_{APC}=S_{ABC}\)
=>\(S_{ABC}=S_{PBC}+\frac12\times S_{PBC}+\frac13\times S_{PBC}=\frac{11}{6}\times S_{BPC}\)
=>\(S_{BPC}=\frac{6}{11}\times S_{ABC}\)
b: Ta có: \(AN=\frac13\times NC\)
=>\(CN=\frac34\times CA\)
=>\(S_{PNC}=\frac34\times S_{PAC}=\frac34\times\frac12\times S_{CPB}=\frac38\times S_{BPC}\)
=>\(\frac{PN}{PB}=\frac38\)