Suppose ABC is a triangle with AB of unit length. D and E are the points lying on AB and AC respectively such that BC and DE are parallel. If the area of triangle ABC is twice the area of triangle ADE, then the length of AD is

In $Delta mathrm{ADE}$ and $Delta mathrm{ABC}$,

$angle mathrm{ADE}=angle mathrm{ABC}$

$angle mathrm{AED}=angle mathrm{ACB}$

$herefore riangle mathrm{ADE} cong Delta mathrm{ABC}$

$herefore frac{Delta mathrm{ABC}}{Delta mathrm{ADE}}=frac{mathrm{AB}^{2}}{mathrm{AD}^{2}}=2$

$Rightarrow frac{1}{mathrm{AD}^{2}}=2 Rightarrow mathrm{AD}^{2}=frac{1}{2}$

$Rightarrow mathrm{AD}=frac{1}{sqrt{2}}$ unit