Giải thích các bước giải:

 a) Ta có:

$\begin{array}{l}
\Delta ABC;\widehat B = {90^0};BA = 3cm;CB = 4cm\\
 \Rightarrow \left\{ \begin{array}{l}
AC = \sqrt {A{B^2} + B{C^2}}  = 5cm\\
BH = \dfrac{{2{S_{ABC}}}}{{AC}} = \dfrac{{BA.BC}}{{AC}} = 2,4cm
\end{array} \right.
\end{array}$

Vậy $AC = 5cm;BH = 2,4cm$

b) Ta có;

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {BIH} = \widehat {BHC} = {90^0}\\
\widehat Bchung
\end{array} \right.\\
 \Rightarrow \Delta BIH \sim \Delta BHC\left( {g.g} \right)\\
 \Rightarrow \dfrac{{BH}}{{BC}} = \dfrac{{BI}}{{BH}}\\
 \Rightarrow B{H^2} = BI.BC
\end{array}$

c) Ta có:

$\widehat {BIH} = \widehat {BKH} = \widehat {KBI} = {90^0}$

$\to $ Tứ giác $BIHK$ là hình chữ nhật.

$\to BH=KI$

d) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {BIE} = \widehat {FKB} = {90^0}\\
\widehat {IBE} = \widehat {KFB}\left( {BI//HF} \right)
\end{array} \right.\\
 \Rightarrow \Delta BIE \sim \Delta FKB\left( {g.g} \right)\\
 \Rightarrow \dfrac{{IE}}{{KB}} = \dfrac{{BI}}{{FK}}\\
 \Rightarrow FK.IE = BI.BK\left( 1 \right)
\end{array}$

Và:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {CIH} = \widehat {HKA} = {90^0}\\
\widehat {CHI} = \widehat {HAK}\left( {HI//AB} \right)
\end{array} \right.\\
 \Rightarrow \Delta CHI \sim \Delta HAK\left( {g.g} \right)\\
 \Rightarrow \dfrac{{CI}}{{HK}} = \dfrac{{HI}}{{AK}}\\
 \Rightarrow CI.AK = HI.HK\left( 2 \right)
\end{array}$

Từ $\left( 1 \right),\left( 2 \right) \Rightarrow FK.IE = CI.AK\left( {do:HI = BK;BI = HK} \right)$

$ \Rightarrow \dfrac{{CI}}{{FK}} = \dfrac{{IE}}{{KA}}$