Prove that BD/AB=(2-sqrt of 2)/2

Question

D and E are points on the sides AB and AC respectively of a triangle ABC such that DE is parallel to BC and divides triangle ABC into two parts equal in area. Prove that BD/AB=(2-sq.rt of 2)/2

Answer 1

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 = AD²/AB²  (Areas of similar triangle) --------(2)

∴ From equation 1 and 2 we get

=> 1/2 = AD²/AB²

=> 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

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