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: MA=MB

=>M là trung điểm của AB

=>\(AM=\frac12\times AB\)

=>\(S_{AMC}=\frac12\times S_{ABC}=\frac12\times20=10\left(\operatorname{cm}^2\right)\)

b: Ta có: MA=MB

=>\(S_{CMA}=S_{CMB};S_{IMA}=S_{IMB}\)

=>\(S_{CMA}-S_{IMA}=S_{CMB}-S_{IMB}\)

=>\(S_{CIA}=S_{CIB}\)

c: Ta có: AN=2NC

=>\(S_{BNA}=2\times S_{BNC};S_{INA}=2\times S_{INC}\)

=>\(S_{BNA}-S_{INA}=2\times\left(S_{BNC}-S_{INC}\right)\)

=>\(S_{BIA}=2\times S_{BIC}\)

=>\(S_{AIB}=2\times S_{AIC}\)

TA có: P nằm giữa B và C

=>\(\frac{S_{ABP}}{S_{ACP}}=\frac{BP}{CP};\frac{S_{IPB}}{S_{IPC}}=\frac{PB}{PC}\)

=>\(\frac{S_{ABP}-S_{IBP}}{S_{ACP}-S_{ICP}}=\frac{BP}{CP}\)

=>\(\frac{BP}{CP}=\frac{S_{AIB}}{S_{AIC}}=2\)

=>BP=2CP

Ta có: \(BN=\frac13NC\)

=>\(S_{ANB}=\frac13\times S_{ANC};S_{INB}=\frac13\times S_{INC}\)

=>\(S_{ANB}-S_{INB}=\frac13\times\left(S_{ANC}-S_{INC}\right)\)

=>\(S_{AIB}=\frac13\times S_{AIC}\)

Ta có: MA=MB

=>\(S_{CMA}=S_{CMB};S_{IMA}=S_{IMB}\)

=>\(S_{CMA}-S_{IMA}=S_{CMB}-S_{IMB}\)

=>\(S_{CIA}=S_{CIB}\)

=>\(S_{AIB}=\frac13\times S_{BIC}\)

\(S_{AIC}+S_{AIB}+S_{BIC}=S_{ABC}\)

=>\(S_{ABC}=S_{BIC}+S_{BIC}+\frac13\times S_{BIC}=\frac73\times S_{BIC}\)

=>\(S_{BIC}=280:\frac73=280\times\frac37=120\left(\operatorname{cm}^2\right)\)

=>\(S_{CIA}=S_{CIB}=120\left(\operatorname{cm}^2\right);S_{AIB}=\frac13\times120=40\left(\operatorname{cm}^2\right)\)

\(BN=\frac13NC\)

=>\(BN=\frac14\times BC\)

=>\(S_{BIN}=\frac14\times S_{BIC}=\frac14\times120=30\left(\operatorname{cm}^2\right)\)

\(BM=\frac12BA\)

=>\(S_{BIM}=\frac12\times S_{BIA}=\frac12\times40=20\left(\operatorname{cm}^2\right)\)

\(S_{BMIN}=S_{BMI}+S_{BNI}\)

\(=30+20=50\left(\operatorname{cm}^2\right)\)

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Kakashi _kun nói dối đó

a , S^AMC = 2/3 S^ABC

S^AMC  là : 60 : 3 x 2 = 40 [ cm² ]

b , S^CMN = 1/2 S^AMC

    S^BMN = 1/2 S^ABM 

S^CMN = 40 : 2 = 20 [ cm² ]

S^BMN = [ 60 - 40 ] : 2 = 10 [ cm² ]

S^BNC = 10 + 20 = 30 [ cm²  ]

Đáp số : a , 40 cm²

               b , 30 cm²

b30

A 40