Đáp án:
a) Theo Pytago ta có:
$\begin{array}{l}
B{C^2} = A{B^2} + A{C^2} = {12^2} + {9^2} = 225\\
\Rightarrow BC = 15\left( {cm} \right)\\
\Rightarrow DC + DB = 15\left( {cm} \right)\\
Theo\,t/c:\\
\frac{{DC}}{{AC}} = \frac{{DB}}{{AB}}\\
\Rightarrow \frac{{DC}}{9} = \frac{{DB}}{{12}}\\
\Rightarrow \frac{{DC}}{{DB}} = \frac{3}{4}
\end{array}$
b)
Xét ΔAHB và ΔCAB có:
+ góc AHB = góc CAB = 90 độ
+ góc ABH chung
=>ΔAHB ~ ΔCAB (g-g)
$\begin{array}{l}
\Rightarrow \frac{{AH}}{{AC}} = \frac{{AB}}{{BC}} = \frac{{12}}{{15}} = \frac{4}{5}\\
\Rightarrow AH = \frac{4}{5}.12 = \frac{{48}}{5}\left( {cm} \right)
\end{array}$