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

Answer: Option B

Solution

Answer: Option B

Solution:

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