Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\Delta ADB;\widehat {ADB} = {90^0}\\
\Rightarrow \sin A = \dfrac{{BD}}{{AB}}\left( 1 \right)
\end{array}$
Lại có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {BEH} = \widehat {BDA} = {90^0}\\
\widehat Bchung
\end{array} \right.\\
\Rightarrow \Delta BEH \sim \Delta BDA\left( {g.g} \right)\\
\Rightarrow \dfrac{{BE}}{{BD}} = \dfrac{{BH}}{{BA}}\\
\Rightarrow \dfrac{{BE}}{{BH}} = \dfrac{{BD}}{{BA}}
\end{array}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat Bchung\\
\dfrac{{BE}}{{BH}} = \dfrac{{BD}}{{BA}}
\end{array} \right.\\
\Rightarrow \Delta BED \sim \Delta BHA\left( {g.g} \right)\\
\Rightarrow \dfrac{{ED}}{{HA}} = \dfrac{{BD}}{{BA}}\left( 2 \right)\\
\left( 1 \right),\left( 2 \right) \Rightarrow \dfrac{{ED}}{{HA}} = \sin A
\end{array}$
Ta có đpcm.
