AM+MB=AB

=>MB=4-3=1(cm)

Xét ΔABC có MN//BC

nên \(\frac{AM}{MB}=\frac{AN}{NC}\)

=>\(\frac{2}{NC}=\frac31=3\)

=>\(NC=\frac23\) (cm)

Xét ΔABC có MN//BC

nên \(\frac{MN}{BC}=\frac{AM}{AB}\)

=>\(\frac{MN}{6}=\frac34\)

=>\(MN=6\cdot\frac34=\frac{18}{4}=4,5\left(\operatorname{cm}\right)\)

a) Do MN//BC nên theo hệ quả của ĐL Ta-let ta có \(\dfrac{AM}{AB}\)=\(\dfrac{MN}{BC}\)

\(\Rightarrow\) \(\dfrac{2}{4}\) = \(\dfrac{MN}{6}\)\(\Rightarrow\) MN = \(\dfrac{2\times6}{4}\)\(\Rightarrow\) MN = 3 cm

b) Do MN//BC nên theo ĐL Ta-let ta có \(\dfrac{AM}{AB}\)=\(\dfrac{AN}{AC}\)

\(\Rightarrow\)\(\dfrac{12}{15}\)=\(\dfrac{AN}{18}\)\(\Rightarrow\) AN = \(\dfrac{12\times18}{15}\) = 14,4 cm

Xét ΔABC có MN//BC

nên \(\frac{AM}{MB}=\frac{AN}{NC}\)

=>\(\frac{AN}{12}=\frac46=\frac23\)

=>\(AN=12\cdot\frac23=8\left(\operatorname{cm}\right)\)

AC=AN+NC

=8+12

=20(cm)

a: AN=4,2cm

b: MN=2,4cm

ghi rõ cách giải được không ạ

Xét ΔABC có AM/AB=AN/AC

nên MN//BC

BM=AB-AM=2cm

Xét ΔABC có MN//BC

nên AM/MB=AN/NC

=>3/NC=2

hay NC=1,5(cm)

=>CA=4,5(cm)

\(BC=\sqrt{6^2+4.5^2}=7.5\left(cm\right)\)

Xét \(\Delta ABC:MN//BC\left(gt\right).\)

\(\Rightarrow\dfrac{AM}{MB}=\dfrac{AN}{NC}\left(Talet\right).\)

\(\Rightarrow\dfrac{AN}{NC}=\dfrac{1}{3}.\)

\(\Rightarrow\dfrac{AN}{AN+NC}=\dfrac{1}{1+3}.\Leftrightarrow\dfrac{AN}{AC}=\dfrac{1}{4}.\)

AM/MB=AN/NC

=>3/NC=2/3

=>NC=4,5cm