Đáp án:

a) Xét ΔABI và ΔHCI có:

+ góc BAI = góc CHI = 90 độ

+ góc AIB = góc HIC (đối đỉnh)

=> ΔABI ~ ΔHCI (g-g)

b)

Do ΔABI ~ ΔHCI nên góc ABI = góc ICH

Mà BI là phân giác góc ABC
=> góc ABI = góc IBC

=> góc IBC = góc ICH

c)

$\begin{array}{l}
Theo\,Pytago:\\
B{C^2} = A{B^2} + A{C^2} = {6^2} + {8^2} = 100\\
 \Rightarrow BC = 10\left( {cm} \right)\\
Theo\,t.c:\\
\dfrac{{AI}}{{AB}} = \dfrac{{IC}}{{BC}}\\
 \Rightarrow \dfrac{{AI}}{6} = \dfrac{{IC}}{{10}} = \dfrac{{AI + IC}}{{6 + 10}} = \dfrac{8}{{16}} = \dfrac{1}{2}\\
 \Rightarrow \left\{ \begin{array}{l}
AI = 3\left( {cm} \right)\\
IC = 5\left( {cm} \right)
\end{array} \right.\\
d)\Delta ABI;\Delta HBC:\\
 + \widehat {ABI} = \widehat {HBC}\\
 + \widehat {BAI} = \widehat {BHC} = {90^0}\\
 \Rightarrow \Delta ABI \sim \Delta HBC\left( {g - g} \right)\\
 \Rightarrow \dfrac{{{S_{ABI}}}}{{{S_{HBC}}}} = {\left( {\dfrac{{AI}}{{HC}}} \right)^2}\\
Do:\Delta ABI \sim \Delta HCI\\
 \Rightarrow \dfrac{{BI}}{{CI}} = \dfrac{{AB}}{{HC}}\\
\left( {BI = \sqrt {A{B^2} + A{I^2}}  = \sqrt {{6^2} + {3^2}}  = 3\sqrt 5 } \right)\\
 \Rightarrow \dfrac{{3\sqrt 5 }}{5} = \dfrac{6}{{HC}}\\
 \Rightarrow HC = \dfrac{{10}}{{\sqrt 5 }}\\
 \Rightarrow \dfrac{{AI}}{{HC}} = \dfrac{3}{{\dfrac{{10}}{{\sqrt 5 }}}} = \dfrac{{3\sqrt 5 }}{{10}}\\
 \Rightarrow \dfrac{{{S_{ABI}}}}{{{S_{HBC}}}} = {\left( {\dfrac{{3\sqrt 5 }}{{10}}} \right)^2} = \dfrac{{45}}{{100}} = \dfrac{9}{{20}}
\end{array}$