Asked in: March 2026 board paperIn \(\Delta ABC\), A – X – B and A – Y – C such that segment XY || side BC. Segment XY divides \(\Delta ABC\) in two equal areas, then find \(\dfrac{BX}{AB}\).

Answer: \(\dfrac{BX}{AB} = \dfrac{\sqrt{2} - 1}{\sqrt{2}} = \dfrac{2 - \sqrt{2}}{2}\)

[Figure: \(\Delta ABC\) with X on AB and Y on AC, XY parallel to BC; \(\Delta AXY\) and trapezium XBCY have equal areas.]

XY || BC, so \(\angle AXY \cong \angle ABC\) (corresponding angles) and \(\angle A\) is common. \(\therefore \Delta AXY \sim \Delta ABC\) (AA test).

XY divides the triangle into two equal parts, so \(A(\Delta AXY) = \dfrac{1}{2} A(\Delta ABC)\), i.e. \(\dfrac{A(\Delta AXY)}{A(\Delta ABC)} = \dfrac{1}{2}\).

By the theorem of areas of similar triangles, \(\dfrac{AX^2}{AB^2} = \dfrac{1}{2}\) \(\therefore \dfrac{AX}{AB} = \dfrac{1}{\sqrt{2}}\).

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

Rationalising, \(\dfrac{BX}{AB} = \dfrac{2 - \sqrt{2}}{2}\) (about 0.29).