Đáp án:
Vì AIHK là hình chữ nhật( 3 góc vuông)
=> Ik = AH
Xét ΔABC vuông tại A có
$\begin{array}{l}
A{B^2} + A{C^2} = B{C^2}(Pytago)\\
hay{3^2} + {4^2} = B{C^2}\\
\Rightarrow BC = 5\\
Lại có:\\
{S_{\Delta ABC}} = \frac{1}{2}AB.AC\\
{S_{\Delta ABC}} = \frac{1}{2}AH.BC\\
\Rightarrow AB.AC = AH.BC\\
hay3.4 = AH.5\\
\Rightarrow 12 = AH.5\\
\Rightarrow AH = \frac{{12}}{5}\\
\Rightarrow IK = \frac{{12}}{5}
\end{array}$
$\begin{array}{l}
b)Xet:\Delta HBI;\Delta HAK:\\
+ \widehat {BHI} = \widehat {AHK}\left( { + \widehat {IHA} = {{90}^0}} \right)\\
+ \widehat {HIB} = \widehat {HKA} = {90^0}\\
\Rightarrow \Delta HBI \sim \Delta HAK\left( {g - g} \right)\\
\Rightarrow \frac{{HB}}{{HA}} = \frac{{HI}}{{HK}} = \frac{{BI}}{{AK}}\\
\Rightarrow HK.HB = HI.HA\\
c)\Delta ABC;\Delta HAC:\\
+ \widehat {ACB}\,chung\\
+ \widehat {BAC} = \widehat {AHC} = {90^0}\\
\Rightarrow \Delta ABC \sim \Delta HAC\left( {g - g} \right)\\
\Rightarrow \frac{{AB}}{{AH}} = \frac{{AC}}{{CH}}\\
\Rightarrow AB.CH = AH.AC
\end{array}$
