a)
Xét $\Delta AEC$ và $\Delta ADB$, ta có:
$\widehat{BAC}$ chung, $\widehat{AEC}=\widehat{ADB}=90{}^\circ $
$\Rightarrow \Delta AEC\backsim\Delta ADB\left( g.g \right)$
$\Rightarrow \dfrac{AE}{AD}=\dfrac{AC}{AB}\Rightarrow AE.AB=AD.AC$
b)
Xét $\Delta ADE$ và $\Delta ABC$, ta có:
$\widehat{BAC}$ chung, $\dfrac{AD}{AB}=\dfrac{AE}{AC}$ (vì $\dfrac{AE}{AD}=\dfrac{AC}{AB}$)
$\Rightarrow \Delta ADE\backsim\Delta ABC\left( c.g.c \right)$
$\Rightarrow \dfrac{{{S}_{\Delta ADE}}}{{{S}_{\Delta ABC}}}={{\left( \dfrac{AD}{AB} \right)}^{2}}={{\cos }^{2}}A$
$\Rightarrow {{S}_{\Delta ADE}}={{S}_{\Delta ABC}}.{{\cos }^{2}}A$
c)
$\dfrac{AD}{AB}=\cos \widehat{BAC}=\cos 60{}^\circ =\dfrac{1}{2}$
Vì $\Delta ADE\backsim\Delta ABC\left( cmt \right)$
$\Rightarrow \dfrac{DE}{BC}=\dfrac{AD}{AB}=\dfrac{1}{2}$
$\Rightarrow BC=2DE=2.5=10cm$
d)
$BD=\sqrt{3}AD\Rightarrow \dfrac{BD}{AD}=\sqrt{3}\Rightarrow \tan \widehat{BAC}=\sqrt{3}\Rightarrow \widehat{BAC}=60{}^\circ $
$\Delta BEC,\Delta BDC$ lần lượt vuông tại $E,D$ có trung tuyến $EF,DF$
$\Rightarrow EF=DF=\dfrac{1}{2}BC$
$\Rightarrow \Delta DEF$ cân tại $F$ $\left( 1 \right)$
$\Delta FDC$ cân tại $F$$\Rightarrow \widehat{CFD}=180{}^\circ -2\widehat{ACB}$
$\Delta FEB$ cân tại $F$ $\Rightarrow \widehat{BFE}=180{}^\circ -2\widehat{ABC}$
$\Rightarrow \widehat{CFD}+\widehat{BFE}=360{}^\circ -2\left( \widehat{ACB}+\widehat{ABC} \right)$
$\Rightarrow \widehat{CFD}+\widehat{BFE}=360{}^\circ -2\left( 180{}^\circ -\widehat{BAC} \right)$
$\Rightarrow \widehat{CFD}+\widehat{BFE}=360{}^\circ -2\left( 180{}^\circ -60{}^\circ \right)$
$\Rightarrow \widehat{CFD}+\widehat{BFE}=120{}^\circ $
$\Rightarrow \widehat{EFD}=60{}^\circ $ $\left( 2 \right)$
Từ $\left( 1 \right)$ và $\left( 2 \right)$ $\Rightarrow \Delta DEF$ là tam giác đều
