Đáp án:
a) Xét ΔHBA và ΔABC có:
+ góc B chung
+ góc BHA = góc BAC = 90 độ
=>ΔHBA ~ ΔABC (g-g)
b) Theo Pytago ta có:
$\begin{array}{l}
B{C^2} = A{B^2} + A{C^2} = {12^2} + {16^2} = 400\\
\Rightarrow BC = 20\left( {cm} \right)\\
Do:\Delta HBA \sim \Delta ABC\\
\Rightarrow \frac{{BH}}{{AB}} = \frac{{AB}}{{BC}} = \frac{{AH}}{{AC}}\\
\Rightarrow \left\{ \begin{array}{l}
\frac{{BH}}{{12}} = \frac{{12}}{{20}} = \frac{3}{5}\\
\frac{{AH}}{{16}} = \frac{3}{5}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
BH = \frac{{36}}{5}\left( {cm} \right)\\
AH = \frac{{48}}{5}\left( {cm} \right)
\end{array} \right.
\end{array}$
