Đáp án:
a) `D` nằm trên cạnh `AB` nên `AD+DB = AB`
`=> AD=8-2=6` (cm)
$\begin{array}{l}
\Rightarrow \dfrac{{AE}}{{AD}} = \dfrac{9}{6} = \dfrac{3}{2}\\
\dfrac{{AD}}{{AC}} = \dfrac{6}{{12}} = \dfrac{1}{2}
\end{array}$
b) Ta có:
$\begin{array}{l}
\dfrac{{AD}}{{AB}} = \dfrac{6}{8} = \dfrac{3}{4}\\
\dfrac{{AE}}{{AC}} = \dfrac{9}{{12}} = \dfrac{3}{4}\\
\Rightarrow \dfrac{{AD}}{{AB}} = \dfrac{{AE}}{{AC}}\\
Xet:\Delta ADE;\Delta ABC:\\
+ \dfrac{{AD}}{{AB}} = \dfrac{{AE}}{{AC}}\\
+ \widehat A\text{ chung}\\
\Rightarrow \Delta ADE \sim \Delta ABC\left( {c - g - c} \right)
\end{array}$
c) Do:
$\begin{array}{l}
\dfrac{{AD}}{{AB}} = \dfrac{{AE}}{{AC}}\\
\Rightarrow \dfrac{{AD}}{{AE}} = \dfrac{{AB}}{{AC}}
\end{array}$
Theo tính chất đường phân giác trong tam giác:
$\begin{array}{l}
\dfrac{{IB}}{{IC}} = \dfrac{{AB}}{{AC}}\\
\Rightarrow \dfrac{{IB}}{{IC}} = \dfrac{{AD}}{{AE}}\\
\Rightarrow IB.AE = IC.AD
\end{array}$
