Đáp án:
a) Ta có:
$\begin{array}{l}
\frac{{AN}}{{AM}} = \frac{{AB}}{{AC}}\left( {\frac{4}{5} = \frac{{12}}{{15}}} \right)\\
\Rightarrow \Delta ANB \sim \Delta AMC\left( {c - c} \right)
\end{array}$
b) DO ΔABN ~ ΔACM
=> góc ABN = góc ACM
Xét ΔOMB và ΔONC có:
+ góc OBM = góc OCN
+ góc BOM = góc CON (đối đỉnh)
=> ΔOMB ~ ΔONC (cg-g)
$\begin{array}{l}
\Rightarrow \frac{{OM}}{{OB}} = \frac{{ON}}{{OC}}\\
\Rightarrow OM.OC = ON.OB
\end{array}$
c) Ta có:
$\begin{array}{l}
\frac{{AM}}{{MB}}.\frac{{BH}}{{HC}}.\frac{{CN}}{{AN}} = 1\\
\Rightarrow \frac{5}{7}.\frac{{BH}}{{HC}}.\frac{{11}}{4} = 1\\
\Rightarrow \frac{{BH}}{{HC}} = \frac{{28}}{{55}}
\end{array}$
