a) Áp dụng hệ thức lượng, ta được:
$AD.AB = AH^2$
$AE.AC = AH^2$
$\Rightarrow AD.AB = AE.AC$
b) Ta có: $AB.AC = BC.AH = 2S_{ABC} \Rightarrow BC = \dfrac{AB.AC}{AH}$
$\star \, DB.CE.BC$
$=DB.CE.\dfrac{AB.AC}{AH}$
$=\dfrac{(DB.AB).(CE.AC)}{AH}$
$=\dfrac{BH^2.CH^2}{AH}$
$=\dfrac{AH^4}{AH} = AH^3$
c) Ta có: $ΔABH \sim ΔCAH$
$\Rightarrow \dfrac{AB}{AC} = \dfrac{BH}{AH}$
$\Rightarrow \left(\dfrac{AB}{AC}\right)^2 = \dfrac{BH^2}{AH^2} = \dfrac{BH^2}{BH.CH} = \dfrac{BH}{CH}$
$\Rightarrow \left(\dfrac{AB}{AC}\right)^4 = \left(\dfrac{BH}{CH}\right)^2 =\dfrac{BD.AB}{CE.AC}$
$\Rightarrow \left(\dfrac{AB}{AC}\right)^3 = \dfrac{BD}{CE}$
d) Ta có: $ΔADH \sim ΔCEH$
$\Rightarrow \dfrac{HD}{HE} = \dfrac{AD}{CE}$
$\Rightarrow \dfrac{AB}{AC}\cdot\dfrac{HD}{HE}=\dfrac{AD.AB}{CE.AC}= \dfrac{AH^2}{CH^2} = \dfrac{BH.CH}{CH^2} = \dfrac{BH}{CH} = \dfrac{AB^2}{AC^2}$
$\Rightarrow \dfrac{HD}{HE} = \dfrac{AB}{AC}$
$\Rightarrow \left(\dfrac{HD}{HE}\right)^3 = \left(\dfrac{AB}{AC}\right)^3 = \dfrac{DB}{EC}$
