Question

The line segment XY is parallel to side AC of triangle ABC and it divides the triangle into two parts of equal areas. Find the ratio AX/AB.

Answer 1

by using midpoint theorem AX/XB is equal to 1/1

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

XY || AC

∠BXY = ∠BAC (Corresponding ∠)

∠BYX = ∠BCA (Corresponding ∠)

$\tt{\rightarrow\triangle BXY\sim\triangle BAC.. (By\;AA}$

★Now

${\dfrac{ar(\triangle BAC)}{ar(\triangle BXY)}=(\dfrac{AB}{BX})^{2}}$

★Given

$\tt{\rightarrow ar(\triangle BXY)=\dfrac{1}{2}ar(\triangle BAC)}$

$\tt{\rightarrow\dfrac{BAC}{1/2(BAC)}=(\dfrac{AB}{BX})^{2}}$

$\tt{\rightarrow(\dfrac{AB}{BX})^{2}=\dfrac{2}{1}}$

$\tt{\rightarrow\dfrac{AB}{BX}=\dfrac{\sqrt{2}}{1}}$

$\tt{\rightarrow\dfrac{BX}{AB}=\dfrac{1}{\sqrt{2}}}$

$\tt{\rightarrow\dfrac{AB-AX}{AB}=\dfrac{1}{\sqrt{2}}}$

$\tt{\rightarrow\dfrac{AX}{AB}=1-\dfrac{1}{\sqrt{2}}}$

$\tt{\rightarrow\dfrac{AX}{AB}=\dfrac{\sqrt{2}-1}{\sqrt{2}}}$

$\tt{\rightarrow\dfrac{AX}{AB}=\dfrac{\sqrt{2}(\sqrt{2}-1)}{2}}$

$\tt{\rightarrow\dfrac{AX}{AB}=\dfrac{2-\sqrt{2}}{2}}$