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)

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

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: AN=4,2cm

b: MN=2,4cm

ghi rõ cách giải được không ạ

a) Ta có: \(\dfrac{AN}{AB}=\dfrac{3}{6}=\dfrac{1}{2}\)

\(\dfrac{AM}{AC}=\dfrac{4.5}{9}=\dfrac{1}{2}\)

Do đó: \(\dfrac{AN}{AB}=\dfrac{AM}{AC}\)\(\left(=\dfrac{1}{2}\right)\)

Xét ΔANM và ΔABC có 

\(\dfrac{AN}{AB}=\dfrac{AM}{AC}\)(cmt)

\(\widehat{BAC}\) chung

Do đó: ΔANM\(\sim\)ΔABC(c-g-c)

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

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

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

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

Xét ΔABC có MN//BC

nên AM/MB=AN/NC

=>4/MB=2/8=1/4

=>MB=16cm

Xét ΔAMN và ΔACB có

AM/AC=AN/AB

góc A chung

Do đo: ΔAMN∼ΔACB

Xét \(\Delta AMN\) và \(\Delta ACB\) :

\(\left\{{}\begin{matrix}\widehat{MAN}=\widehat{BAC}\left(g-c\right)\\\dfrac{AM}{AC}=\dfrac{AN}{AB}\end{matrix}\right.\)

\(\Rightarrow\Delta AMN\sim\Delta ACB\left(c-g-c\right)\)

A B C M N 3 5 10 16 6

Ta có: \(\frac{MB}{AB}=\frac{MB}{AM+MB}=\frac{5}{8}\)

\(\frac{NC}{AC}=\frac{10}{16}=\frac{5}{8}\)

=> \(\frac{MB}{AB}=\frac{NC}{AC}\)Theo định lí Ta-lét đảo

=>  MN // BC

Áp dụng hệ quả định lí Ta-lét vào \(\Delta ABC\)có MN // BC

=> \(\frac{MN}{BC}=\frac{AM}{AB}\)=> \(BC=\frac{MN.AB}{AM}=\frac{8.6}{3}=16\)