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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"`.
Solution
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)`.
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