Giải thích các bước giải:

 a) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHB} = \widehat {CHA} = {90^0}\\
\widehat {HAB} = \widehat {HCA}\left( { + \widehat {ABC} = {{90}^0}} \right)
\end{array} \right.\\
 \Rightarrow \Delta AHB \sim \Delta CHA\left( {g.g} \right)\\
 \Rightarrow \dfrac{{AH}}{{CH}} = \dfrac{{HB}}{{HA}}\\
 \Rightarrow A{H^2} = BH.CH
\end{array}$

b) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AEF} = \widehat {BEH}\left( {dd} \right)\\
\widehat {AFE} = \widehat {BHE} = {90^0}
\end{array} \right.\\
 \Rightarrow \Delta AEF \sim \Delta BEH\left( {g.g} \right)
\end{array}$

c) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHC} = \widehat {BAC} = {90^0}\\
\widehat Cchung
\end{array} \right.\\
 \Rightarrow \Delta AHC \sim \Delta BAC\left( {g.g} \right)\\
 \Rightarrow \dfrac{{AH}}{{BA}} = \dfrac{{AC}}{{BC}}\\
 \Rightarrow \dfrac{{AH}}{{AC}} = \dfrac{{BA}}{{BC}}\left( 1 \right)
\end{array}$

Lại có:

$AD$ là phân giác $\Delta AHC$$ \Rightarrow \dfrac{{DH}}{{DC}} = \dfrac{{AH}}{{AC}}\left( 2 \right)$

Và:

$BK$ là phân giác $\Delta ABC$ $ \Rightarrow \dfrac{{KA}}{{KC}} = \dfrac{{BA}}{{BC}}\left( 3 \right)$

Từ $\left( 1 \right),\left( 2 \right),\left( 3 \right) \Rightarrow \dfrac{{DH}}{{DC}} = \dfrac{{KA}}{{KC}}$

$ \Rightarrow KD//AH$

d) Ta có:

Do $EH//KD$ nên theo ĐL Talet ta có: $\dfrac{{EH}}{{KD}} = \dfrac{{BH}}{{BD}}\left( 4 \right)$

Lại có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {BDK} = \widehat {BAK} = {90^0}\left( {do:KD//AH} \right)\\
BKchung\\
\widehat {DBK} = \widehat {ABK} = \dfrac{1}{2}\widehat {ABC}
\end{array} \right.\\
 \Rightarrow \Delta BDK = \Delta BAK\left( {ch - gn} \right)\\
 \Rightarrow BD = BA\\
 \Rightarrow \dfrac{{BH}}{{BD}} = \dfrac{{BH}}{{BA}}\left( 5 \right)
\end{array}$

Và:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHB} = \widehat {CAB} = {90^0}\\
\widehat Bchung
\end{array} \right.\\
 \Rightarrow AHB \sim \Delta CAB\left( {g.g} \right)\\
 \Rightarrow \dfrac{{AB}}{{CB}} = \dfrac{{HB}}{{AB}}\left( 6 \right)
\end{array}$

Từ $\left( 4 \right),\left( 5 \right),\left( 6 \right) \Rightarrow \dfrac{{EH}}{{KD}} = \dfrac{{AB}}{{CB}}$

$ \Rightarrow \dfrac{{EH}}{{AB}} = \dfrac{{KD}}{{BC}}$