**Solution:**

**Given:** ABC is a triangle with DE || BC divides the triangle into two parts equal in areas.

**To find: **BD/AB=(2-sq.rt of 2)/2

Area of ΔADE = Area of Trapezium BDEC (As given)

∴ Area of ΔADE = 1/2 area of ΔABC ------------(1)

In ΔADE and ΔABC

∠ADE = ∠ABC (Corresponding angles)

∠AED = ∠ACB (Corresponding angles)

ΔADE ∼ ΔABC (AA similarity)

∴ Area of ΔADE / Area of ΔABC = AD^{2}/AB^{2 }(Areas of similar triangle) --------(2)

∴ From equation 1 and 2 we get

=> 1/2 = AD^{2}/AB^{2}

=> AD/AB = 1/√ 2

∴ AB – AD = √ 2 AD – AD

∴ BD = (√ 2 – 1)AD

∴ BD/AB = (√ 2 – 1)AD / √ 2 AD

∴ BD/AB = (√ 2 – 1) / √ 2 --------------(3)

Multiplying N^{r} and D^{r} of equation 3 by √ 2 , we get

**∴ BD/AB = (2 – √ 2) / 2**