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