Ta có: AM+MB=AB

=>\(MB=AB-AM=AB-\frac35\times AB=\frac25\times AB\)

=>\(AM=\frac32\times MB\)

=>\(S_{CMA}=\frac32\times S_{CMB};S_{OMA}=\frac32\times S_{OMB}\)

=>\(S_{CMA}-S_{OMA}=\frac32\times\left(S_{CMB}-S_{OMB}\right)\)

=>\(S_{COA}=\frac32\times S_{COB}\)

Ta có AN+NC=AC

=>NC=AC-AN=1/5AC

=>AN=4NC

=>\(S_{BNA}=4\times S_{BNC};S_{ONA}=4\times S_{ONC}\)

=>\(S_{BNA}-S_{ONA}=4\times\left(S_{BNC}-S_{ONC}\right)\)

=>\(S_{BOA}=4\times S_{BOC}\)

=>\(\frac{S_{AOB}}{S_{AOC}}=4:\frac32=4\times\frac23=\frac83\)

Vì AM=MB

nên \(S_{CMA}=S_{CMB};S_{OMA}=S_{OMB}\)

=>\(S_{CMA}-S_{OMA}=S_{CMB}-S_{OMB}\)

=>\(S_{COA}=S_{COB}\)

Ta có: AN+NC=AC

=>\(NC=AC-AN=AC-\frac34\times AC=\frac14\times AC\)

=>\(AN=3\times NC\)

=>\(S_{BNA}=3\times S_{BNC};S_{ONA}=3\times S_{ONC}\)

=>\(S_{BNA}-S_{ONA}=3\times\left(S_{BNC}-S_{ONC}\right)\)

=>\(S_{BOA}=3\times S_{BOC}\)

=>\(S_{BOA}=3\times S_{COA}\)

=>\(\frac{S_{AOB}}{S_{AOC}}=3\)

a: Ta có: \(AN=\frac13\times AC\)

=>\(S_{BNA}=\frac13\times S_{ABC}\) (1)

Ta có: \(AM=\frac13\times AB\)

=>\(S_{AMC}=\frac13\times S_{ABC}\) (2)

Từ (1),(2) suy ra \(S_{BNA}=S_{AMC}=\frac13\times S_{ABC}\)

b: Ta có: \(AM=\frac13\times AB\)

=>\(S_{AMN}=\frac13\times S_{ANB}=\frac13\times\frac13\times S_{ABC}=\frac19\times S_{ABC}\)

Ta có: \(S_{AMN}+S_{BMNC}=S_{ABC}\)

=>\(S_{BMNC}=S_{ABC}-\frac19\times S_{ABC}=\frac89\times S_{ABC}=\frac89\times36=32\left(\operatorname{cm}^2\right)\)

bạn vào rồi cóp lại hộ mình với

a: Ta có: AN+NC=AC

=>\(NC=AC-AN=AC-\frac14\times AC=\frac34\times AC\)

=>\(S_{BNC}=\frac34\times S_{ABC}=\frac34\times64=48\left(\operatorname{cm}^2\operatorname{}^{}\right)\)

b: Ta có: \(AN=\frac14\times AC\)

=>\(S_{ABN}=\frac14\times S_{ABC}\)

Ta có: \(AM=\frac14\times AB\)

=>\(S_{AMN}=\frac14\times S_{ANB}=\frac14\times\frac14\times S_{ABC}=\frac{1}{16}\times S_{ABC}\)

=>\(\frac{S_{AMN}}{S_{ABC}}=\frac{1}{16}\)

NA=NC

=>\(S_{BNA}=S_{BNC};S_{ONA}=S_{ONC}\)

=>\(S_{BNA}-S_{ONA}=S_{BNC}-S_{ONC}\)

=>\(S_{BOA}=S_{BOC}\) (1)

Ta có: AM+MB=AB

=>\(MB=AB-AM=AB-\frac13\times AB=\frac23\times AB\)

=>MB=2xAM

=>\(S_{CMB}=2\times S_{CMA};S_{OMB}=2\times S_{OMA}\)

=>\(S_{CMB}-S_{OMB}=2\times\left(S_{CMA}-S_{OMA}\right)\)

=>\(S_{COB}=2\times S_{COA}\) (2)

Từ (1),(2) suy ra \(S_{BOA}=2\times S_{COA}\)

=>\(S_{AOB}=2\times S_{AOC}\)

Vì D nằm giữa B và C nên ta có:

\(\frac{S_{ADB}}{S_{ADC}}=\frac{DB}{DC};\frac{S_{ODB}}{S_{ODC}}=\frac{DB}{DC}\)

=>\(\frac{DB}{DC}=\frac{S_{ABD}-S_{OBD}}{S_{ACD}-S_{OCD}}\)

=>\(\frac{DB}{DC}=\frac{S_{ABO}}{S_{ACO}}=2\)