Ta có: $DE//BC \, (gt)$
$\Rightarrow ΔADE \sim ΔABC \, (Theo \,\, Thales)$
$\Rightarrow \dfrac{S_{ADE}}{S_{ABC}} = \left(\dfrac{AD}{AB}\right)^2$
mà $AD = AK \, (gt)$
$\Rightarrow \dfrac{AD}{AB} = \dfrac{AK}{AB} = \dfrac{1}{\sqrt{2}}$
Do đó:
$\dfrac{S_{ADE}}{S_{ABC}} = \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{1}{2}$
Vậy $S_{ADE} = \dfrac{1}{2}S_{ABC}$