Question

In $△ABC,D$ is a point on AB and E is a point on BC such that DE || AC and $ar\left(DBE\right)=\frac{1}{2}ar\left(ABC\right)$. Find $\frac{AD}{AB}$

• A. $\frac{1-\sqrt{2}}{2}$
• B. $\frac{\sqrt{2}-1}{\sqrt{2}}$
• C. $\frac{\sqrt{2}-1}{2}$
• D. $\frac{\sqrt{2}+1}{2}$

Answer 1

The correct option is B $\frac{\sqrt{2}-1}{\sqrt{2}}$

In $△ABC$

$DE\parallel BC$

$\frac{AD}{BD}=\frac{AE}{EC}$ Basic proportionality theorm

$AD=\frac{BD}{2}$

$\frac{AD}{BD}=\frac{1}{3}$

$\frac{area\left(DBE\right)}{area\left(ABC\right)}={\left(\frac{AD}{AB}\right)}^2$

$\sqrt{\frac{area\left(DBE\right)}{area\left(ABC\right)}}=\left(\frac{AD}{AB}\right)$

$1-\frac{AD}{AB}$

$=1-\frac{1}{\sqrt{2}}$

$=\frac{\sqrt{2}-1}{\sqrt{2}}$