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Question
In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC
If AD = 8x – 7, DB = 5x – 3, AE = 4x – 3 and EC = 3x – 1, find the value of x
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Solution
Given AD = 8x – 7; BD = 5x – 3; AE = 4x – 3; EC = 3x – 1
In ∆ABC we have DE || BC
By Basic proportionality theorem
`"AD"/"DB" = "AE"/"EC"`
`(8x - 7)/(5x - 3) = (4x - 3)/(3x - 1)`
(8x – 7) (3x – 1) = (4x – 3) (5x – 3)
24x2 – 8x – 21x + 7 = 20x2 – 12x – 15x + 9
24x2 – 20x2 – 29x + 27x + 7 – 9 = 0
4x2 – 2x – 2 = 0
2x2 – x – 1 = 0 ...(Divided by 2)
2x2 – 2x + x – 1 = 0
2x(x – 1) + 1 (x – 1) = 0
(x – 1) (2x + 1) = 0
x – 1 = 0 or 2x + 1 = 0
x = 1 or 2x = – 1 ⇒ x = `– 1/2` ...(Negative value will be omitted)
The value of x = 1
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