a)
Xét $\Delta AEB$ và $\Delta AFC$, có:
$\widehat{BAC}$ chung, $\widehat{AEB}=\widehat{AFC}=90{}^\circ $
$\Rightarrow \Delta AEB\backsim\Delta AFC\left( g.g \right)$
$\Rightarrow \dfrac{AE}{AF}=\dfrac{AB}{AC}\Rightarrow AB.AF=AC.AE$
b)
Xét $\Delta AEF$ và $\Delta ABC$, có:
$\widehat{BAC}$ chung,$\dfrac{AE}{AB}=\dfrac{AF}{AC}$ (vì $\dfrac{AE}{AF}=\dfrac{AB}{AC}$)
$\Rightarrow \Delta AEF\backsim\Delta ABC\left( c.g.c \right)$
c)
Vì $\Delta AEF\backsim\Delta ABC$
$\Rightarrow \widehat{AEF}=\widehat{ABC}$
Mà: $\begin{cases}\widehat{AEF}+\widehat{BEF}=90^\circ\\\widehat{ABC}+\widehat{BCF}=90^\circ\end{cases}$
Nên: $\widehat{BEF}=\widehat{BCF}$
d)
Xét $\Delta CEB$ và $\Delta CDA$, có:
$\widehat{ACB}$ chung, $\widehat{CEB}=\widehat{CDA}=90{}^\circ $
$\Rightarrow \Delta CEB\backsim\Delta CDA\left( g.g \right)$
$\Rightarrow \dfrac{CE}{CD}=\dfrac{CB}{CA}$$\Rightarrow \dfrac{CE}{CB}=\dfrac{CD}{CA}$
Xét $\Delta CED$ và $\Delta CBA$, có:
$\widehat{ACB}$ chung, $\dfrac{CE}{CB}=\dfrac{CD}{CA}\left( cmt \right)$
$\Rightarrow \Delta CED\backsim\Delta CBA\left( c.g.c \right)$
$\Rightarrow \widehat{CED}=\widehat{CBA}$
Mà: $\begin{cases}\widehat{CED}+\widehat{BED}=90^\circ\\\widehat{CBA}+\widehat{BCF}=90^\circ\end{cases}$
Nên: $\widehat{BED}=\widehat{BCF}$
Kết hợp ý c, ta được $\widehat{BEF}=\widehat{BED}$
$\Rightarrow EH$ là phân giác $\widehat{DEF}$
