a) DE//BC nên:
$\frac{AD}{AB}=$ $\frac{AE}{AC}⇔$ $\frac{AD}{AE}$ = $\frac{AB}{AC}$
b) DE//BC nên:
$\frac{BD}{AB}=$ $\frac{EC}{AC}⇔BD.AC=CE.AB$
c) DK//DH ( cùng ⊥AC) nên:
$\frac{DK}{BH}=$ $\frac{AD}{AB}=$ $\frac{DE}{BC}⇔$ $\frac{DK}{DE}=$ $\frac{BH}{BC}$
d) Ta có:
$S_{ADE}=\frac{DK.AE}{2}$
$S_{ABC}=\frac{BH.AC}{2}$
Suy ra: $\frac{S_{ADE}}{S_{ABC}}=$ $\frac{DK.AE}{BH.AC}=$$\frac{DK}{BH}.$ $\frac{AE}{AC}=$ $\frac{AD}{AB}.$ $\frac{DE}{BC}$