3.
a)
Xét $\Delta AEB$ và $\Delta AFC$, ta có:
$\widehat{BAC}$ chung, $\widehat{AEB}=\widehat{AFC}=90{}^\circ $
$\Rightarrow \Delta AEB\backsim\Delta AFC\left( g.g \right)$
$\Rightarrow \dfrac{AE}{AB}=\dfrac{AF}{AC}$
Xét $\Delta AEF$ và $\Delta ABC$, ta có:
$\widehat{BAC}$ chung, $\dfrac{AE}{AB}=\dfrac{AF}{AC}\left( cmt \right)$
$\Rightarrow \Delta AEF\backsim\Delta ABC\left( c.g.c \right)$
$\Rightarrow \dfrac{FE}{BC}=\dfrac{AE}{AB}=\cos A$
$\Rightarrow FE=BC.\cos A=2a.\cos A$
b)
Chứng minh tương tự câu a, ta được 2 ý;
$\Delta BFD\backsim\Delta BCA\left( c.g.c \right)\Rightarrow \dfrac{FD}{AC}=\dfrac{BF}{BC}=\cos B$
$\Delta CED\backsim\Delta CBA\Rightarrow \dfrac{ED}{AB}=\dfrac{CE}{CB}=\cos C$
$\Rightarrow \dfrac{FE}{BC}\cdot \dfrac{FD}{AC}\cdot \dfrac{ED}{AB}=\cos A.\cos B.\cos C$
c)
Vì $\Delta AEF\backsim\Delta ABC\left( cmt \right)$
$\Rightarrow \dfrac{{{S}_{\Delta AEF}}}{{{S}_{\Delta ABC}}}={{\left( \dfrac{AE}{AB} \right)}^{2}}={{\cos }^{2}}A$
4)
a)
Xét $\Delta ADB$ và $\Delta CDH$, ta có:
$\widehat{DAB}=\widehat{DCH}$ (cùng phụ $\widehat{ABC}$)
$\widehat{ADB}=\widehat{CDH}=90{}^\circ $
$\Rightarrow \Delta ADB\backsim\Delta CDH\left( g.g \right)$
$\Rightarrow \dfrac{AD}{CD}=\dfrac{BD}{HD}\Rightarrow AD.HD=BD.CD$
