Đáp án:
a) Xét ΔABI và ΔHCI có:
+ góc BAI = góc CHI = 90 độ
+ góc BIA = góc CIH = 90 độ
=> ΔABI ~ ΔHCI (g-g)
b)
Do ΔABI ~ ΔHCI (cmt)
=>góc ABI = góc ICH
Mà góc ABI = góc HBI
=> góc HBI = góc ICH
c)
Theo Pytago:
$\begin{array}{l}
A{B^2} + A{C^2} = B{C^2} = {6^2} + {8^2} = 100\\
\Rightarrow BC = 10\left( {cm} \right)\\
Theo\,t/c:\\
\frac{{AI}}{{AB}} = \frac{{IC}}{{BC}}\\
\Rightarrow \frac{{AI}}{6} = \frac{{IC}}{{10}} = \frac{{AI + IC}}{{16}} = \frac{{AC}}{{16}} = \frac{1}{2}\\
\Rightarrow \left\{ \begin{array}{l}
AI = 3\left( {cm} \right)\\
IC = 5\left( {cm} \right)
\end{array} \right.
\end{array}$