Đáp án:
Theo Pytago:
$\begin{array}{*{20}{l}}
{A{B^2} + B{C^2} = A{C^2}}\\
{ \Rightarrow A{C^2} = {9^2} + {{15}^2} = 306}\\
{ \Rightarrow AC = 3\sqrt {34} \left( {cm} \right)}
\end{array}$
Do BI là phân giác của góc ABC
$\begin{array}{l}
\Rightarrow \dfrac{{IC}}{{BC}} = \dfrac{{IA}}{{AB}}\\
= \dfrac{{IC}}{{15}} = \dfrac{{IA}}{9} = \dfrac{{IC + IA}}{{15 + 9}} = \dfrac{{3\sqrt {34} }}{{24}} = \dfrac{{\sqrt {34} }}{8}\\
\Rightarrow \left\{ \begin{array}{l}
IC = \dfrac{{15\sqrt {34} }}{8}\left( {cm} \right)\\
IA = \dfrac{{9\sqrt {34} }}{8}\left( {cm} \right)
\end{array} \right.\\
\Rightarrow \dfrac{{IA}}{{AC}} = \dfrac{{\dfrac{{9\sqrt {34} }}{8}}}{{3\sqrt {34} }} = \dfrac{3}{8}\\
Do:IE//BC\\
\Rightarrow \dfrac{{IE}}{{BC}} = \dfrac{{IA}}{{AC}} = \dfrac{3}{8}\\
\Rightarrow IE = \dfrac{3}{8}.15 = \dfrac{{45}}{8}
\end{array}$
