Giải thích các bước giải:

 Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AND} = \widehat {ADC} = {90^0}\\
\widehat Achung
\end{array} \right.\\
 \Rightarrow \Delta AND \sim \Delta ADC\left( {g.g} \right)\\
 \Rightarrow \dfrac{{AN}}{{AD}} = \dfrac{{AD}}{{AC}}\\
 \Rightarrow AN.AC = A{D^2}
\end{array}$

Chứng minh tương tự có: $\Delta ADM \sim \Delta ABD\left( {g.g} \right)$

$\begin{array}{l}
 \Rightarrow AM.AB = A{D^2}\\
 \Rightarrow AM.AB = AN.AC\\
 \Rightarrow \dfrac{{AM}}{{AC}} = \dfrac{{AN}}{{AB}}
\end{array}$

Khi đó:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat Achung\\
\dfrac{{AM}}{{AC}} = \dfrac{{AN}}{{AB}}
\end{array} \right.\\
 \Rightarrow \Delta AMN \sim \Delta ACB\left( {g.g} \right)
\end{array}$

Lại có:

$\dfrac{{AM}}{{AC}} = \dfrac{{AN}}{{AB}} \Rightarrow \dfrac{{AM}}{{AN}} = \dfrac{{AC}}{{AB}}\left( 1 \right)$

Mặt khác:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat Achung\\
\widehat {AFC} = \widehat {AEB} = {90^0}
\end{array} \right.\\
 \Rightarrow \Delta AFC \sim \Delta AEB\left( {g.g} \right)\\
 \Rightarrow \dfrac{{AF}}{{AE}} = \dfrac{{AC}}{{AB}}\left( 2 \right)
\end{array}$

Từ $\left( 1 \right),\left( 2 \right) \Rightarrow \dfrac{{AF}}{{AE}} = \dfrac{{AM}}{{AN}}$

$ \Rightarrow MN//EF$