In a Δ ABC, D and E are points on the sides AB and AC respectively su

1.	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.
Answer» 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

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.

Answer»

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