a: Ta có: \(AN=2\times NC\)

=>\(S_{PNA}=2\times S_{PNC};S_{MNA}=2\times S_{MNC}\)

=>\(S_{PNA}-S_{MNA}=2\times\left(S_{PNC}-S_{MNC}\right)\)

=>\(S_{PMA}=2\times S_{PMC}\)

Ta có: \(CM=3\times MB\)

=>\(S_{PMC}=3\times S_{PMB}\)

=>\(S_{PMA}=2\times3\times S_{PMB}=6\times S_{PMB}\)

=>PA=6BP

PB+BA=PA

=>BA=PA-PB=6BP-BP=5BP

=>5BP=6

=>BP=1,2(cm)

PA=PB+BA=1,2+6=7,2(cm)

b: Ta có: \(BM+MC=BC\)

=>BC=3MB+MB=4MB

=>\(S_{ABC}=4\times S_{AMB}\)

TA có: AN+NC=AC

=>AC=2NC+NC=3NC

=>\(AN=\frac23\times AC\)

=>\(S_{AMN}=\frac23\times S_{AMC}\)

Ta có: CM=3/4CB

=>\(S_{AMC}=\frac34\times S_{ABC}\)

=>\(S_{AMN}=\frac34\times\frac23\times S_{ABC}=\frac12\times S_{ABC}\)

PA=1,2AB

=>\(S_{PMA}=\frac65\times S_{BAM}=\frac65\times\frac14\times S_{ABC}=\frac{3}{10}\times S_{ABC}\)

=>\(\frac{S_{PMA}}{S_{AMN}}=\frac{3}{10}:\frac12=\frac{3}{10}\times2=\frac35\)

=>MP=3/5MN

a: Ta có: \(AN=2\times NC\)

=>\(S_{PNA}=2\times S_{PNC};S_{MNA}=2\times S_{MNC}\)

=>\(S_{PNA}-S_{MNA}=2\times\left(S_{PNC}-S_{MNC}\right)\)

=>\(S_{PMA}=2\times S_{PMC}\)

Ta có: \(CM=3\times MB\)

=>\(S_{PMC}=3\times S_{PMB}\)

=>\(S_{PMA}=2\times3\times S_{PMB}=6\times S_{PMB}\)

=>PA=6BP

PB+BA=PA

=>BA=PA-PB=6BP-BP=5BP

=>5BP=6

=>BP=1,2(cm)

PA=PB+BA=1,2+6=7,2(cm)

b: Ta có: \(BM+MC=BC\)

=>BC=3MB+MB=4MB

=>\(S_{ABC}=4\times S_{AMB}\)

TA có: AN+NC=AC

=>AC=2NC+NC=3NC

=>\(AN=\frac23\times AC\)

=>\(S_{AMN}=\frac23\times S_{AMC}\)

Ta có: CM=3/4CB

=>\(S_{AMC}=\frac34\times S_{ABC}\)

=>\(S_{AMN}=\frac34\times\frac23\times S_{ABC}=\frac12\times S_{ABC}\)

PA=1,2AB

=>\(S_{PMA}=\frac65\times S_{BAM}=\frac65\times\frac14\times S_{ABC}=\frac{3}{10}\times S_{ABC}\)

=>\(\frac{S_{PMA}}{S_{AMN}}=\frac{3}{10}:\frac12=\frac{3}{10}\times2=\frac35\)

=>MP=3/5MN

a: Xét ΔABC có AM/AB=AN/AC
nên MN//BC

=>\(\dfrac{S_{AMN}}{S_{ABC}}=\left(\dfrac{AM}{AB}\right)^2=\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)

b: Xét ΔABE có MK//BE

nên AK/AE=AM/AB=2/3

=>AK=2/3AE