Question

In the figure, XY $\parallel$ AC and XY divides the triangular region ABC into two equal areas.

Determine AX : XB.

• A. $(\sqrt{3}-1):1$
• B. $(\sqrt{2}-1):1$
• C. $(\sqrt{7}-1):1$
• D. $(\sqrt{1}1-1):1$

Answer 1

The correct option is B $(\sqrt{2}-1):1$

Ar. AXYC $=$ Ar. BXY

Ar. AXYC +Ar. BXY $=2$ Ar. BXY

Ar.ABC $=2$ Ar. BXY

Hence, $\frac{Ar.ABC}{Ar.BXY}=2$ ...(I)

In $△$s ABC and BXY,

$\angle XBY=\angle ABC$ (Common angle)

$\angle BXY=\angle BAC$ (Corresponding angles of parallel lines XY and AC)

$\angle BYX=\angle BCA$ (Corresponding angles of parallel lines XY and AC)

$△ABC\sim △XBY$ (AAA postulate)

Now, $\frac{Ar.ABC}{Ar.BXY}=\frac{A{B}^2}{B{X}^2}$ (Similar triangle Property)

$2=\frac{A{B}^2}{B{X}^2}$

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

$AB=\sqrt{2}BX$

$AX+XB=\sqrt{2}XB$

$AX=(\sqrt{2}-1)XB$

$AX:XB=(\sqrt{2}-1):1$