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In a Δ ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1, and CE = 5x – 3, find the value of x.

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Given: AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3 

Required to find x. 

By using Thales Theorem, [As DE ∥ BC] 

AD/BD = AE/CE 

So, (4x–3)/ (3x–1) = (8x–7)/ (5x–3) 

(4x – 3)(5x – 3) = (3x – 1)(8x – 7) 

4x(5x – 3) -3(5x – 3) = 3x(8x – 7) -1(8x – 7) 

20x2 – 12x – 15x + 9 = 24x2 – 29x + 7 

20x2 - 27x + 9 = 24x2 - 29x + 7 

⇒ -4x2 + 2x + 2 = 0 

4x2 – 2x – 2 = 0 

4x2 – 4x + 2x – 2 = 0 

4x(x – 1) + 2(x – 1) = 0 

(4x + 2)(x – 1) = 0 

⇒ x = 1 or x = -\(\frac{2}{4}\) 

We know that the side of triangle can never be negative. Therefore, we take the positive value. 

∴ x = 1

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