In ABC, XY is parallel to AC and divides the triangle into the two parts of equal area Then AX AB is

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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 AX / AB.

Solution

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, ar (ABC)/ar (XBY) = \((AB/XB)^2\)

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

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

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

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

or, 1 – $\frac{XB}{AB}$ = 1 – $\frac{1}{\sqrt{2}}$ or, $\frac{AB-XB}{AB}$ = ($\sqrt{2}$–1)/$\sqrt{2}$, i.e., $\frac{AX}{AB}$ =(2–$\sqrt{2}$)/2