Giải thích các bước giải:

a) Ta có:

$\begin{array}{l}
\Delta DHE;\widehat {DHE} = {90^0};DE = 5cm;DH = 4cm\\
 \Rightarrow EH = \sqrt {D{E^2} - D{H^2}}  = 3cm
\end{array}$

b) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {DHE} = \widehat {DKF} = {90^0}\\
DE = DF\\
\widehat Dchung
\end{array} \right.\\
 \Rightarrow \Delta DHE = \Delta DKF\left( {ch - gn} \right)
\end{array}$

c) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {EKF} = \widehat {FHE} = {90^0}\\
EFchung\\
\widehat {KEF} = \widehat {HFE}\left( {\widehat {DEF} = \widehat {DFE}} \right)
\end{array} \right.\\
 \Rightarrow \Delta EKF = \Delta FHE\left( {ch - gn} \right)\\
 \Rightarrow \widehat {EFK} = \widehat {FEH}\\
 \Rightarrow \widehat {OFE} = \widehat {OEF}
\end{array}$

$ \Rightarrow \Delta OEF$ cân ở $O$

d) Ta có:

$\begin{array}{l}
\Delta EKF = \Delta FHE\left( {ch - gn} \right)\\
 \Rightarrow EK = FH
\end{array}$

Khi đó:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {EKO} = \widehat {FHO} = {90^0}\\
OE = OF\\
EK = FH
\end{array} \right.\\
 \Rightarrow \Delta EKO = \Delta FHO\left( {ch - cgv} \right)\\
 \Rightarrow OK = OH
\end{array}$

$ \Rightarrow \Delta OHK$ cân ở $O$

$\begin{array}{l}
 \Rightarrow \widehat {OKH} = \dfrac{{{{180}^0} - \widehat {KOH}}}{2}\\
 \Rightarrow \widehat {OKH} = \dfrac{{{{180}^0} - \widehat {EOF}}}{2}\\
 \Rightarrow \widehat {OKH} = \widehat {OFE}\\
 \Rightarrow HK//EF
\end{array}$

e) Ta có:

$DI = DE = DF$

$ \Rightarrow \Delta DIE$ cân ở $D$

$ \Rightarrow \widehat {DIE} = \widehat {IED}$

Xét $\Delta IEF$ có:

$\begin{array}{l}
\widehat {IEF} + \widehat {EIF} + \widehat {IFE} = {180^0}\\
 \Rightarrow \widehat {IEF} + \widehat {IED} + \widehat {DEF} = {180^0}\\
 \Rightarrow 2\widehat {IEF} = {180^0}\\
 \Rightarrow \widehat {IEF} = {90^0}\\
 \Rightarrow IE \bot EF
\end{array}$