mình cũng thấy nó hơi khó vì nó là bài toán chứng minh hình và mình cũng chưa giải được nên mới đi hỏi

a: Ta có: AM+MC=AC

=>\(MC=AC-AM=AC-\frac23\times AC=\frac13\times AC\)

=>\(MC=\frac12\times MA\)

=>\(S_{BMC}=\frac12\times S_{BMA}\)

b: Bổ sung đề: Tính \(S_{ABC}\)

Ta có: BN+NM=BM

=>\(NM=BM-BN=BM-\frac23\times BM=\frac13\times BM\)

=>\(S_{ANM}=\frac13\times S_{ABM}\)

=>\(S_{ABM}=18:\frac13=54\left(\operatorname{cm}^2\right)\)

Ta có: \(AM=\frac23\times AC\)

=>\(S_{ABM}=\frac23\times S_{ABC}\)

=>\(S_{ABC}=54:\frac23=54\times\frac32=81\left(\operatorname{cm}^2\right)\)

a: \(AM=\frac23MC\)

=>\(MC=\frac32AM\)

AM+MC=AC

=>\(AC=AM+\frac32AM=\frac52AM\)

=>\(AM=\frac25AC\)

=>\(S_{ABM}=\frac25\times S_{ABC}=\frac25\times60=24\left(\operatorname{cm}^2\right)\)

b: Ta có: BN+NC=BC

=>\(BC=BN+\frac13BN=\frac43BN\)

=>\(CN=\frac14CB;BN=\frac34BC\)

TA có: \(CM=\frac35\times CA\)

=>\(S_{BMC}=\frac35\times S_{ABC}\)

\(CN=\frac14CB\)

=>\(S_{MNC}=\frac14\times S_{BMC}=\frac14\times\frac35\times S_{ABC}=\frac{3}{20}\times S_{ABC}\)

TA có: \(S_{AMNB}+S_{MNC}=S_{ABC}\)

=>\(S_{AMNB}=S_{ABC}-\frac{3}{20}\times S_{ABC}=\frac{17}{20}\times S_{ABC}=\frac{17}{20}\times60=51\left(\operatorname{cm}^2\right)\)

c: Ta có: \(AM=\frac23\times MC\)

=>\(S_{BMA}=\frac23\times S_{BMC};S_{OMA}=\frac23\times S_{OMC}\)

=>\(S_{BMA}-S_{OMA}=\frac23\times\left(S_{BMC}-S_{OMC}\right)\)

=>\(S_{BOA}=\frac23\times S_{BOC}\)

=>\(S_{BOC}=\frac32\times S_{BOA}\)

Ta có: \(NC=\frac13\times NB\)

=>\(S_{ANC}=\frac13\times S_{ANB};S_{ONC}=\frac13\times S_{ONB}\)

=>\(S_{ANC}-S_{ONC}=\frac13\times\left(S_{ANB}-S_{ONB}\right)\)

=>\(S_{AOC}=\frac13\times S_{AOB}=\frac29\times S_{BOC}\)

Ta có: \(S_{BOC}+S_{AOC}+S_{AOB}=S_{ABC}\)

=>\(S_{ABC}=S_{OAB}+\frac32\times S_{OAB}+\frac13\times S_{OAB}=S_{OAB}\left(\frac43+\frac32\right)=S_{AOB}\cdot\frac{11}{6}\)

=>\(S_{AOB}=\frac{6}{11}\cdot S_{ABC}=\frac{6}{11}\cdot60=\frac{360}{11}\left(\operatorname{cm}^2\right)\)

=>\(\frac{S_{AOB}}{S_{ABN}}=\frac{\frac{6}{11}\cdot S_{ABC}}{\frac34\cdot S_{ABC}}=\frac{6}{11}:\frac34=\frac{6}{11}\times\frac43=\frac{24}{33}=\frac{8}{11}\)

=>\(\frac{AO}{AN}=\frac{8}{11}\)

=>\(\frac{AO}{ON}=\frac83\)

Là 90 nha mình ko biết cách làm