Problem Statement
In $\triangle ABC$, $D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $AD = 8x - 7, DB = 5x - 3, AE = 4x - 3$ and $EC = (3x - 1)$. Find the value of $x$ if $DE \parallel BC$.
Verified Solution & Marking Scheme
Apply Basic Proportionality Theorem
Since $DE \parallel BC$:
$$\frac{AD}{DB} = \frac{AE}{EC}$$
$$\frac{8x - 7}{5x - 3} = \frac{4x - 3}{3x - 1}$$
Cross Multiply and Expand
$$(8x - 7)(3x - 1) = (4x - 3)(5x - 3)$$
$$24x^2 - 8x - 21x + 7 = 20x^2 - 12x - 15x + 9$$
$$24x^2 - 29x + 7 = 20x^2 - 27x + 9$$
$$4x^2 - 2x - 2 = 0 \implies 2x^2 - x - 1 = 0$$
Factorize and Reject Invalid Roots
$$2x^2 - 2x + x - 1 = 0 \implies 2x(x - 1) + 1(x - 1) = 0$$
$$(2x + 1)(x - 1) = 0$$
If $x = -1/2$, side length $AE = 4(-1/2) - 3 = -5 < 0$ (length cannot be negative).
Therefore $x = 1$.