In Δ ABC, A-X-B and A-Y-C is such that segment XY || side BC. Segment XY divides Δ ABC in two equal areas, then find \(\dfrac{BX}{AB}\).

Answer: \(\dfrac{BX}{AB} = \dfrac{2 - \sqrt{2}}{2}\) (≈ 0.293)

XY || BC, so Δ AXY ~ Δ ABC (AA test).

Let \(\dfrac{AX}{AB} = k\). Then \(\dfrac{A(\Delta AXY)}{A(\Delta ABC)} = k^2\) (areas of similar triangles).

XY divides the triangle into two equal areas, so A(Δ AXY) = \(\dfrac{1}{2}\) A(Δ ABC), giving \(k^2 = \dfrac{1}{2}\), i.e. \(k = \dfrac{1}{\sqrt{2}}\).

∴ \(\dfrac{AX}{AB} = \dfrac{1}{\sqrt{2}}\), so \(\dfrac{BX}{AB} = 1 - \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2} - 1}{\sqrt{2}} = \dfrac{2 - \sqrt{2}}{2}\).