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Question
In ΔABC, D and E are the mid-point on AB and AC such that DE || BC.
If AD = 4x - 3, AE = 8x - 7, BD = 3x - 1 and CE = 5x - 3,Find x.
Solution
In ΔADE and ΔABC
∠D = ∠B and ∠C = ∠E ...(DE || BC)
⇒ ΔADE ∼ ΔABC
∴ `"AD"/"DB" = "AE"/"EC"`
⇒ `(4x - 3)/(3x - 1) = (8x - 7)/(5x - 3)`
⇒ (4x - 3) x (5x - 3) = (8x - 7) x (3x - 1)
⇒ 20x2 - 15x - 12x + 9 = 24x2 - 21x - 8x + 7
⇒ 20x2 - 27x + 9 = 24x2 29x + 7
⇒ 4x2 - 2x - 2 = 0
⇒ `x(x - 1) + (1)/(2)(x - 1)` = 0
⇒ `(x + 1/2)` = 0; x - 1 = 0
⇒ x = `-(1)/(2); x = 1`
∴ x = 1.
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