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Cho tam giác ABC nhọn, 3 đường cao AD, BE, CF cắt nhau tại H. c) Từ D vẽ DM vuông góc AB, DN vuông góc AC. CmTam giác AMN đồ dạng vói ACB Cm: MN//EF.
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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$