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Kẻ ND//AB (D thuộc AB).
Có: \(MC=\dfrac{1}{2}AM;MC+AM=AC\)
\(\Rightarrow\dfrac{AM}{AC}=\dfrac{2}{3};\dfrac{MC}{AC}=\dfrac{1}{3}\).
Có: \(NC=2BN;NC+BN=BC\)
\(\Rightarrow\dfrac{NC}{BC}=\dfrac{2}{3};\dfrac{BN}{BC}=\dfrac{1}{3}\)
△ABC có: ND//AB.
\(\Rightarrow\dfrac{ND}{AB}=\dfrac{DC}{AB}=\dfrac{2}{3}\) (định lí Ta-let)
\(\Rightarrow ND=\dfrac{2}{3}AB=\dfrac{2}{3}.6=4\left(cm\right)\).
\(\dfrac{AD}{AC}=\dfrac{BN}{BC}=\dfrac{1}{3}=\dfrac{MC}{AC}\Rightarrow AD=MC=\dfrac{1}{3}AC\)
Mà \(AD+DM+MC=AC\Rightarrow AD=DM=MC=\dfrac{1}{3}AC\); \(AM=DC=\dfrac{2}{3}AC\).
\(\Rightarrow\dfrac{MD}{AM}=\dfrac{1}{2}\)
△APM có: DN//AP.
\(\Rightarrow\dfrac{ND}{AP}=\dfrac{MD}{AM}=\dfrac{1}{2}\) (hệ quả định lí Ta-let)
\(\Rightarrow AP=2ND=2.4=8\left(cm\right)\)
a: AN=NC
=>\(S_{BNA}=S_{BNC}\)
Ta có: MA=2MB
=>\(S_{CMA}=2\cdot S_{CMB}\)
b: Ta có: AM+MB=AB
=>AB=2MB+MB=3MB
=>\(S_{AGB}=3\cdot S_{MBG}=3\cdot5=15\left(\operatorname{cm}^2\right)\)
Vì NA=NC
nên \(S_{BNA}=S_{BNC};S_{GNA}=S_{GNC}\)
=>\(S_{BNA}-S_{GNA}=S_{BNC}-S_{GNC}\)
=>\(S_{BGA}=S_{BGC}\)
=>\(S_{BGC}=15\left(\operatorname{cm}^2\right)\)
Ta có: AM=2MB
=>\(S_{CMA}=2\cdot S_{CMB};S_{GMA}=2\cdot S_{GMB}\)
=>\(S_{CMA}-S_{GMA}=2\cdot\left(S_{CMB}-S_{GMB}\right)\)
=>\(S_{CGA}=2\cdot S_{CGB}=2\cdot15=30\left(\operatorname{cm}^2\right)\)
\(S_{ABC}=S_{AGB}+S_{AGC}+S_{BGC}\)
\(=15+15+30=60\left(\operatorname{cm}^2\right)\)
Ta có: AM=MB
=>\(S_{CMA}=S_{CMB};S_{KMA}=S_{KMB}\)
=>\(S_{CMA}-S_{KMA}=S_{CMB}-S_{KMB}\)
=>\(S_{CKA}=S_{CKB}\left(1\right)\)
Ta có: AN=NC
=>\(S_{BNA}=S_{BNC};S_{KNA}=S_{KNC}\)
=>\(S_{BNA}-S_{KNA}=S_{BNC}-S_{KNC}\)
=>\(S_{BKA}=S_{BKC}\left(2\right)\)
Từ (1),(2) suy ra \(S_{CKA}=S_{AKB}=S_{BKC}\)
mà \(S_{CKA}+S_{AKB}+S_{BKC}=S_{ABC}\)
nên \(S_{BKC}=\frac{S_{ABC}}{3}=\frac{91}{3}\left(\operatorname{cm}^2\right)\)