b)
Xét $\Delta DHB$ và $\Delta DCA$, ta có:
$\widehat{HDB}=\widehat{CDA}=90{}^\circ $
$\widehat{DBH}=\widehat{DAC}$ ( cùng phụ $\widehat{BCA}$ )
$\to \Delta DHB\sim \Delta DCA\,\,\left( \,g.g\, \right)$
$\to \dfrac{DH}{DC}=\dfrac{DB}{DA}$
$\to DH.DA=DB.DC$
c)
Xét $\Delta CAD$ và $\Delta CBE$, ta có:
$\widehat{C}$ chung
$\widehat{CDA}=\widehat{CEB}=90{}^\circ $
$\to \Delta CAD\sim \Delta CBE$
$\to \dfrac{CA}{CB}=\dfrac{CD}{CE}$
$\to \dfrac{CE}{CB}=\dfrac{CD}{CA}$
Xét $\Delta CED$ và $\Delta CBA$, ta có:
$\widehat{C}$ chung
$\dfrac{CE}{CB}=\dfrac{CD}{CA}$ ( cmt )
$\to \Delta CED\sim \Delta CBA$
$\to \widehat{CED}=\widehat{CBA}$
Mà:
$\begin{cases}\widehat{CED}+\widehat{BED}=90{}^\circ\\\widehat{CBA}+\widehat{BAD}=90{}^\circ\end{cases}$
$\to \widehat{BAD}=\widehat{BED}$
