a)
$\Delta ABC$ có hai đường cao $BE,CF$ cắt nhau tại $H$
$\Rightarrow H$ là trực tâm $\Delta ABC$ $\Rightarrow AH\bot BC$ tại $D$
b)
Có: $\Delta AFC\backsim\Delta AEB\left( g.g \right)$
$\Rightarrow \dfrac{AF}{AC}=\dfrac{AE}{AB}$
$\Rightarrow \Delta AFE\backsim\Delta ACB\left( c.g.c \right)$
c)
Vì $\Delta AFE\backsim\Delta ACB$
$\Rightarrow \dfrac{{{S}_{\Delta AFE}}}{{{S}_{\Delta ACB}}}={{\left( \dfrac{AE}{AB} \right)}^{2}}={{\cos }^{2}}A$
$\Rightarrow {{S}_{\Delta AFE}}={{S}_{\Delta ABC}}.{{\cos }^{2}}A$
Có ${{S}_{BFEC}}={{S}_{\Delta ABC}}-{{S}_{\Delta AFE}}$
$\Rightarrow {{S}_{BFEC}}={{S}_{\Delta ABC}}-{{S}_{\Delta ABC}}.{{\cos }^{2}}A$
$\Rightarrow {{S}_{BFEC}}={{S}_{\Delta ABC}}\left( 1-{{\cos }^{2}}A \right)$
$\Rightarrow {{S}_{BFEC}}={{S}_{\Delta ABC}}.{{\sin }^{2}}A$
d)
Gọi $M$ là trung điểm $BC$
Với $HG//BC$ nên theo Ta-let, có:
$\dfrac{AG}{AM}=\dfrac{AH}{AD}=\dfrac{2}{3}\Rightarrow \dfrac{DA}{DH}=3$
Có: $\Delta DBA\backsim\Delta DHC\left( g.g \right)$
$\Rightarrow DB.DC=DH.DA$
Có $\tan B.\tan C=\dfrac{DA}{DB}\cdot \dfrac{DA}{DC}$
$\Rightarrow \tan B.\tan C=\dfrac{D{{A}^{2}}}{DH.DA}$
$\Rightarrow \tan B.\tan C=\dfrac{DA}{DH}=3$
