Question

The line segment XY is parallel to side AC of $△$ABC and it divides the triangle into two parts of equal areas. Find the ratio $\frac{BX}{AB}$.

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{4}{\sqrt{1}}$
• D. $\frac{\sqrt{2}}{1}$

Answer 1

The correct option is A

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

We have XY || AC (Given)

So, $\angle$ BXY $=\angle$ A and $\angle$ BYX $=\angle$ C [Corresponding angles]

Therefore, $△$ ABC $\sim$ $△$ XBY [AA similarity criterion]

So, $\frac{\text{Ar}\left(△ABC\right)}{\text{Ar}\left(△XBY\right)}={\left(\frac{AB}{XB}\right)}^2$ ​

Also, $\text{Ar}\left(△ABC\right)=2\text{Ar}\left(△XBY\right)$

So, $\frac{\text{Ar}\left(△ABC\right)}{\text{Ar}\left(△XBY\right)}=\frac{2}{1}$

Therefore, ${\left(\frac{AB}{XB}\right)}^2=\frac{2}{1}$, i.e., $\frac{AB}{XB}=\frac{\sqrt{2}}{1}$

$\Rightarrow \frac{XB}{AB}=\frac{1}{\sqrt{2}}$