In the given figure, the line segment XY is parallel to side AC of ΔABC and…

Concept: For two similar triangles whose corresponding sides are in ratio k, the ratio of their areas is k2 — area scales with the square of the linear scale factor, not the factor itself.

Application: In ΔABC, XY ∥ AC, so ΔBXY ~ ΔBAC by AA similarity (∠B is common to both triangles, and ∠BXY = ∠BAC as corresponding angles cut by the parallel lines). Let k = BX/BA = BY/BC = XY/AC be their similarity ratio.

1. XY splits ΔABC into two equal-area parts — ΔBXY and trapezium AXYC — so Area(ΔBXY) = (1/2)·Area(ΔABC).

2. By the concept above, Area(ΔBXY)/Area(ΔABC) = k2, so k2 = 1/2.

3. Taking the square root, k = 1/√2 — this is the ratio BX/BA.

4. Since AX = AB − BX, dividing throughout by AB gives AX/AB = 1 − BX/AB = 1 − 1/√2.

5. Simplifying, AX/AB = (√2 − 1)/√2, i.e. AX : AB = (√2 − 1) : √2.

Cross-check: Rationalising (√2 − 1)/√2 by multiplying the numerator and denominator by √2 gives (2 − √2)/2 ≈ 0.293 — the same ratio in another form. Numerically, taking AB = 1 gives BX ≈ 0.707 and AX ≈ 0.293, so Area(ΔBXY)/Area(ΔABC) = (BX/BA)2 ≈ 0.7072 = 0.5 — exactly the equal-area condition given, confirming the result.