Đáp án:
$\begin{array}{l}
a)Theo\,Pytago:\\
B{C^2} = A{B^2} + A{C^2} = {8^2} + {15^2} = 289\\
\Rightarrow BC = 17\left( {cm} \right)\\
Theo\,t/c:\\
\dfrac{{AD}}{{AB}} = \dfrac{{DC}}{{BC}}\\
\Rightarrow \dfrac{{AD}}{8} = \dfrac{{DC}}{{17}} = \dfrac{{AD + DC}}{{8 + 17}} = \dfrac{{AC}}{{25}} = \dfrac{{15}}{{25}} = \dfrac{3}{5}\\
\Rightarrow \left\{ \begin{array}{l}
AD = \dfrac{{24}}{5}\left( {cm} \right)\\
DC = \dfrac{{51}}{5}\left( {cm} \right)
\end{array} \right.\\
b)Xét\,\Delta HBA;\Delta ABC:\\
+ \widehat {HBA}\,chung\\
+ \widehat {BHA} = \widehat {BAC} = {90^0}\\
\Rightarrow \Delta HBA \sim \Delta ABC\left( {g - g} \right)\\
\Rightarrow k = \dfrac{{AB}}{{BC}} = \dfrac{8}{{17}}
\end{array}$
c) Xét ΔABD và ΔCBK có:
+ góc BAD = góc BCK = 90 độ
+ góc ABD = góc CBK (gt)
=> ΔABD ~ ΔCBK (g-g)
$\begin{array}{l}
\Rightarrow \dfrac{{AB}}{{BC}} = \dfrac{{BD}}{{BK}}\\
\Rightarrow AB.BK = BD.BC
\end{array}$
d)
$\begin{array}{l}
Do:\Delta ABD \sim \Delta CBK\\
\Rightarrow \dfrac{{{S_{CBK}}}}{{{S_{ABD}}}} = {\left( {\dfrac{{AB}}{{BC}}} \right)^2} = {\left( {\dfrac{8}{{17}}} \right)^2} = \dfrac{{64}}{{289}}\\
\Rightarrow {S_{CBK}} = \dfrac{{64}}{{289}}.{S_{ABD}}\\
= \dfrac{{64}}{{289}}.\dfrac{1}{2}.AB.AD\\
= \dfrac{{6144}}{{1445}}\left( {c{m^2}} \right)
\end{array}$
e)
Ta có góc HAB và góc ABI đều là góc nhọn
=> góc AIB > 90 độ
=> tam giác IAB là tam giác tù.
