In ΔABC, D and E are points on AB and AC respectively such that DE || BC and DE divides the ΔABC into two parts of equal areas. Then ratio of AD and BD is

A. 1 : 1

B. 1 : $$\sqrt 2 $$ - 1

C. 1 : $$\sqrt 2 $$

D. 1 : $$\sqrt 2 $$ + 1

Answer: Option B

Solution(By Examveda Team)

ar ADE = ar DEBC
So, ar ΔADE = 1 unit² and ar ABC = 2 unit²
$$\eqalign{ & \frac{{{\text{ar}}\,\Delta ADE}}{{{\text{ar}}\,\Delta ABC}} = \frac{{A{D^2}}}{{A{B^2}}} \cr & \frac{1}{2} = {\left( {\frac{{AD}}{{AB}}} \right)^2} \cr & \frac{1}{{\sqrt 2 }} = \frac{{AD}}{{AB}} \cr & \therefore \frac{{AD}}{{DB}} = \frac{1}{{\sqrt 2 - 1}} \cr & \left( {\therefore DB = AB - AD = \sqrt 2 - 1} \right) \cr & {\text{So,}}\,AD:BD = 1:\sqrt 2 - 1 \cr} $$