Question

The line segment XY is parallel to side AC of Δ 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. $\frac{4}{1}$
• 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 (since corresponding angles are equal)

$\therefore$ $△$ ABC $\sim$ $△$ XBY (AA similarity criterion)

The ratio of the area of two similar triangles are equal to the ratio of the squares of their corresponding sides.

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

Also, $ar\left(△ABC\right)=2\times ar\left(△XBY\right)$

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

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

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

Taking reciprocal on both sides, we get

$\frac{BX}{AB}=\frac{1}{\sqrt{2}}$