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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
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: \(AM=\frac12MB\)
=>\(S_{CMA}=\frac12\times S_{CMB};S_{PMA}=\frac12\times S_{PMB}\)
=>\(S_{CMA}-S_{PMA}=\frac12\times\left(S_{CMB}-S_{PMB}\right)\)
=>\(S_{CPA}=\frac12\times S_{CPB}\)
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: \(AN=\frac13\times NC\)
=>\(CN=\frac34\times CA\)
=>\(S_{PNC}=\frac34\times S_{PAC}=\frac34\times\frac12\times S_{CPB}=\frac38\times S_{BPC}\)
=>\(\frac{PN}{PB}=\frac38\)
Mình giải theo cách lớp 5.
a) Có: \(AN+NC=AC\) mà \(AN=\dfrac{1}{2}NC\)
\(\Rightarrow\dfrac{1}{2}NC+NC=AC\Rightarrow\dfrac{3}{2}NC=AC\Rightarrow NC=\dfrac{2}{3}AC\)
\(2AN=\dfrac{2}{3}AC\Rightarrow AN=\dfrac{2}{3}.\dfrac{1}{2}AC=\dfrac{1}{3}AC\)
\(\dfrac{S_{ABN}}{S_{ABC}}=\dfrac{AN}{AC}=\dfrac{1}{3}\Rightarrow S_{ABN}=\dfrac{1}{3}S_{ABC}\left(1\right)\)
\(\dfrac{S_{ACM}}{S_{ABC}}=\dfrac{AM}{AB}=\dfrac{1}{3}\Rightarrow S_{ACM}=\dfrac{1}{3}S_{ABC}\left(2\right)\)
Từ (1) và (2) suy ra:
\(S_{ABN}=S_{ACM}\)
\(\Rightarrow S_{ABN}-S_{AMON}=S_{ACM}-S_{AMON}\)
\(\Rightarrow S_{MOB}=S_{NOC}\).
b) \(\dfrac{S_{AMC}}{S_{AMN}}=\dfrac{AC}{AN}=3\Rightarrow S_{AMC}=3S_{AMN}=3.4,5=13,5\left(cm^2\right)\)
\(\dfrac{S_{ABC}}{S_{AMN}}=\dfrac{AB}{AM}=3\Rightarrow S_{ABC}=3S_{AMN}=3.13,5=40,5\left(cm^2\right)\)
\(\dfrac{S_{NCB}}{S_{ABC}}=\dfrac{NC}{AC}=\dfrac{2}{3}\Rightarrow S_{NCB}=\dfrac{2}{3}S_{ABC}=\dfrac{2}{3}.40,5=27\left(cm^2\right)\)
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
=>\(S_{CMA}=S_{CMB};S_{IMA}=S_{IMB}\)
=>\(S_{CMA}-S_{IMA}=S_{CMB}-S_{IMB}\)
=>\(S_{CIA}=S_{CIB}\)
=>\(S_{AIB}=\frac13\times S_{BIC}\)
\(S_{AIC}+S_{AIB}+S_{BIC}=S_{ABC}\)
=>\(S_{ABC}=S_{BIC}+S_{BIC}+\frac13\times S_{BIC}=\frac73\times S_{BIC}\)
=>\(S_{BIC}=280:\frac73=280\times\frac37=120\left(\operatorname{cm}^2\right)\)
=>\(S_{CIA}=S_{CIB}=120\left(\operatorname{cm}^2\right);S_{AIB}=\frac13\times120=40\left(\operatorname{cm}^2\right)\)
\(BN=\frac13NC\)
=>\(BN=\frac14\times BC\)
=>\(S_{BIN}=\frac14\times S_{BIC}=\frac14\times120=30\left(\operatorname{cm}^2\right)\)
\(BM=\frac12BA\)
=>\(S_{BIM}=\frac12\times S_{BIA}=\frac12\times40=20\left(\operatorname{cm}^2\right)\)
\(S_{BMIN}=S_{BMI}+S_{BNI}\)
\(=30+20=50\left(\operatorname{cm}^2\right)\)