Đáp án:
a) Theo t.c đường phân giác trong tam giác ta có:
$\begin{array}{l}
\dfrac{{BD}}{{CD}} = \dfrac{{AB}}{{AC}} = \dfrac{9}{{12}} = \dfrac{3}{4}\\
\Rightarrow \dfrac{{BD}}{3} = \dfrac{{CD}}{4} = \dfrac{{BD + CD}}{{3 + 4}} = \dfrac{{BC}}{7}\\
TheoPytago:B{C^2} = A{B^2} + A{C^2}\\
\Rightarrow B{C^2} = {9^2} + {12^2} = 225\\
\Rightarrow BC = 15\left( {cm} \right)\\
\Rightarrow \dfrac{{BD}}{3} = \dfrac{{CD}}{4} = \dfrac{{15}}{7}\\
\Rightarrow \left\{ \begin{array}{l}
BD = \dfrac{{45}}{7}\left( {cm} \right)\\
CD = \dfrac{{60}}{7}\left( {cm} \right)
\end{array} \right.\\
b)Xet:\Delta ABC;\Delta EDC:\\
+ \widehat {BAC} = \widehat {DEC} = {90^0}\\
+ \widehat C\,chung\\
\Rightarrow \Delta ABC \sim \Delta EDC\left( {g - g} \right)\\
c)\Delta ABC \sim \Delta EDC\\
\Rightarrow \dfrac{{AB}}{{ED}} = \dfrac{{BC}}{{DC}}\\
\Rightarrow ED = \dfrac{{AB.DC}}{{BC}} = \dfrac{{9.\dfrac{{60}}{7}}}{{15}} = \dfrac{{36}}{7}\left( {cm} \right)\\
d)\dfrac{{{S_{ABD}}}}{{{S_{ADC}}}} = \dfrac{{\dfrac{1}{2}.{h_A}.BD}}{{\dfrac{1}{2}.{h_A}.DC}} = \dfrac{{BD}}{{DC}} = \dfrac{3}{4}
\end{array}$
