Đáp án:

8) Theo t/c đường phân giác ta có:

$\begin{array}{l}
\frac{{IP}}{{IN}} = \frac{{MP}}{{MN}} \Rightarrow \frac{{IP}}{4} = \frac{5}{6}\\
 \Rightarrow IP = \frac{5}{6}.4 = \frac{{10}}{3}\left( {cm} \right)
\end{array}$

9) Theo Pytago ta có:

$\begin{array}{l}
A{C^2} = B{C^2} - A{B^2} = {15^2} - {9^2} = 144\\
 \Rightarrow AC = 12\left( {cm} \right)\\
a)\frac{{AD}}{{AB}} = \frac{{CD}}{{BC}}\\
 \Rightarrow \frac{{AD}}{9} = \frac{{CD}}{{15}} = \frac{{AC + CD}}{{9 + 15}} = \frac{{12}}{{24}} = \frac{1}{2}\\
 \Rightarrow \left\{ \begin{array}{l}
AD = 4,5\left( {cm} \right)\\
CD = 7,5\left( {cm} \right)
\end{array} \right.\\
b)\frac{{BI}}{{ID}} = \frac{{AB}}{{AD}} = \frac{9}{{4,5}} = 2\\
 \Rightarrow BI > ID
\end{array}$