Question

In $\Delta ABC$, $\overline{XY}$ is parallel to $\overline{AC}$ and divides the triangle into the two parts of equal area. Then the $\frac{AX}{AB}$ equals:

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

Answer 1

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

Given that $XY$ is parallel to $AC$

Area of triangle $BXY=\frac{1}{2}\times BX\times BY\times sinB$

Area of triangle $ABC=\frac{1}{2}\times AB\times BC\times sinB$

We have $\frac{1}{2}\times AB\times BC\times sinB=2\times \frac{1}{2}\times BX\times BY\times sinB$

$\Rightarrow \frac{BX}{BA}\times \frac{BY}{BC}=\frac{1}{2}$

$\Rightarrow \frac{BX}{BA}=\frac{1}{\sqrt{2}}$

Therefore $\frac{AX}{AB}=1-\frac{BX}{AB}=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}$

multiplying numerator and denominator by $\sqrt{2}$

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