Question

D and E are points on the sides AB and AC  of ΔABC such that DE | | BC and divides ΔABC into two parts, equal in area. Find `"BD"/"AB"`.

Answer 1

Area(ΔADE) = area(trapezium BCED)
⇒ Area(ΔADE) + Area(ΔADE)
= Area(trapezium BCED) + Area(ΔADE)
⇒ 2 Area(ΔADE) = Area(ΔABC)
In ΔADE and ΔABC,
∠ADE = ∠B        ...(corresponding angles)      
∠A = ∠A
Therefore, ΔADE ∼ ΔABC

∴ `"area(Δ ADE)"/"area(Δ ABC)" = "AD"^2/"AB"^2`

⇒ `"area(ΔADE)"/(2" x area(ΔADE)") = "AD"^2/"AB"^2`

⇒ `(1)/(2) = ("AD"/"AB")^2`

⇒ `"AD"/"AB" = (1)/sqrt(2)`
⇒ AB = `sqrt(2)"AD"`
⇒ AB = `sqrt(2)("AB - BD")`
⇒ `(sqrt(2) - 1)"AB" = sqrt(2)"BD"`

⇒ `"BD"/"AB" = (sqrt(2) - 1)/(sqrt(2)`

= `(2 - sqrt(2))/(2)`.