Đáp án:

1. Xét ΔAHN và ΔACH có:

+ góc ANH = góc AHC = 90 độ

+ góc HAN chung

=>ΔAHN ~ ΔACH (g-g)

2. Áp dụng Pytago ta có:

$\begin{array}{l}
B{H^2} = A{B^2} - A{H^2} = {15^2} - {12^2} = 81\\
 \Rightarrow BH = 9\left( {cm} \right)\\
 + C{H^2} = A{C^2} - A{H^2} = {13^2} - {12^2} = 25\\
 \Rightarrow CH = 5\left( {cm} \right)\\
 \Rightarrow BC = 9 + 5 = 14\left( {cm} \right)
\end{array}$

3. Do ΔAHN ~ ΔACH 

=> AH / AC = AN/AH

$\begin{array}{l}
 \Rightarrow A{H^2} = AC.AN\\
 \Rightarrow AN = \frac{{A{H^2}}}{{AC}} = \frac{{{{12}^2}}}{{13}} = \frac{{144}}{{13}}\left( {cm} \right)\\
TT:AM = \frac{{A{H^2}}}{{AB}} = \frac{{{{12}^2}}}{{15}} = \frac{{144}}{{15}}\left( {cm} \right)\\
 \Rightarrow \frac{{AN}}{{AB}} = \frac{{144/13}}{{15}} = \frac{{144}}{{13.15}}\\
\frac{{AM}}{{AC}} = \frac{{144/15}}{{13}} = \frac{{144}}{{13.15}}\\
 \Rightarrow \frac{{AN}}{{AB}} = \frac{{AM}}{{AC}}
\end{array}$

Xét ΔAMN và ΔACB có:

+ AM / AC = AN/AB

+ góc A chung

=> ΔAMN~ ΔACB (c-g-c)

4. 

$\begin{array}{l}
\frac{{AN}}{{AB}} = \frac{{AM}}{{AC}} = \frac{{MN}}{{BC}} = \frac{{144}}{{13.15}}\\
 \Rightarrow MN = BC.\frac{{144}}{{13.15}} = \frac{{672}}{{65}}\left( {cm} \right)
\end{array}$