The given figure shows a quadrilateral ABCD in which AB is parallel to CD.

If OA = 3x - 19, OB = x - 3, OC = x - 5 and OD = 3, find the value of x.

As, AB || DC,

∴ Figure ABCD is a trapezium.

We know that,

Diagonals of trapezium divides each other proportionally.

$\Rightarrow \dfrac{OA}{OC} = \dfrac{OB}{OD} \\[1em] \Rightarrow \dfrac{3x - 19}{x - 5} = \dfrac{x - 3}{3} \\[1em] \Rightarrow 3(3x - 19) = (x - 3)(x - 5) \\[1em] \Rightarrow 9x - 57 = x^2 - 5x - 3x + 15 \\[1em] \Rightarrow 9x - 57 = x^2 - 8x + 15 \\[1em] \Rightarrow x^2 - 8x - 9x + 15 + 57 = 0 \\[1em] \Rightarrow x^2 - 17x + 72 = 0 \\[1em] \Rightarrow x^2 - 9x - 8x + 72 = 0 \\[1em] \Rightarrow x(x - 9) - 8(x - 9) = 0 \\[1em] \Rightarrow (x - 8)(x - 9) = 0 \\[1em] \Rightarrow x - 8 = 0 \text{ or } x - 9 = 0 \\[1em] \Rightarrow x = 8 \text{ or } x = 9.$

Hence, x = 8, 9.