Giải thích các bước giải:
a) Ta có:
$\left\{ \begin{array}{l}
\widehat {ADC} = \widehat {BEC} = {90^0}\\
\widehat Cchung
\end{array} \right. \Rightarrow \Delta ADC \sim \Delta BEC\left( {g.g} \right)$
b) Ta có:
$\begin{array}{l}
\Delta ADC \sim \Delta BEC \Rightarrow \widehat {DAC} = \widehat {EBC} \Rightarrow \widehat {EBC} = a\\
+ )\sin a = \sin \widehat {DAC} = \frac{{CD}}{{AC}} = CD;\cos a = \cos \widehat {DAC} = \frac{{AD}}{{AC}} = AD\\
\Rightarrow 2\sin a.\cos a = 2.CD.AD = BC.AD\left( 1 \right)\\
+ )\sin A = \frac{{BE}}{{BA}} = BE = BE.AC\left( 2 \right)\\
+ )BC.AD = BE.AC = 2{S_{ABC}}\left( 3 \right)\\
\left( 1 \right),\left( 2 \right),\left( 3 \right) \Rightarrow \sin A = 2\sin a.\cos a
\end{array}$
