In a triangle ABC, XY drawn parallel to the base BC divides AB such that area AXYA = area XBCYX. (as shown below). Find ratio AX/BX.?

Since XY//BC, ΔAXY ~ ΔABC
AX² : AB² = (Area AXYA) : (Area ABCA)
AX² : AB² = (Area AXYA) : (Area ABCA + Area XBCYX)
AX² : AB² = 1 : 2
AX : AB = 1 : √2

Let AX = k, and then AB = (√2)k
BX = AB - AX = (√2 - 1)k

AX/BX = 1/(√2 - 1)
AX/BX = [1/(√2 - 1)] × [(√2 + 1)/√2 + 1]
AX/BX = (√2 + 1)/[(√2)² - 1²]
AX/BX = 1 + √2