Giải thích các bước giải:
a. Do BM là phân giác góc ABC
\(\begin{array}{l}
\to \left\{ \begin{array}{l}
\frac{{MC}}{{MA}} = \frac{{BC}}{{AB}} = \frac{7}{4} \to 4MC = 7MA\\
MC - MA = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
MC = 7\\
MA = 4
\end{array} \right.
\end{array}\)
b. Do AN là phân giác góc BAC
\(\begin{array}{l}
\to \frac{{NC}}{{NB}} = \frac{{AC}}{{AB}} = \frac{4}{5} \to \frac{{NC}}{{NB + NC}} = \frac{5}{{4 + 5}}\\
\to \frac{{NB}}{{BC}} = \frac{5}{9} \to NC = \frac{{5.BC}}{9} = 10\left( {cm} \right)
\end{array}\)
c. \(VT = \frac{{AC}}{{BC}}.\frac{{AB}}{{AC}}.\frac{{BC}}{{AB}} = 1 = VP\)
d. Xét ΔABN có:
\(\frac{{OA}}{{ON}} = \frac{{AB}}{{BN}} = \frac{3}{2} \to AB = \frac{3}{2}BN \to BN = 8\)
Có: \(\begin{array}{l}
\frac{{AM}}{{MC}} = \frac{{AB}}{{BC}} = \frac{{AB}}{{BN + NC}} = \frac{{12}}{{8 + NC}} = \frac{6}{7}\\
\to NC = 6
\end{array}\)
Xét ΔACN có:
\(\frac{{OA}}{{ON}} = \frac{{AC}}{{NC}} = \frac{3}{2} \to AC = \frac{3}{2}NC = 9\)
