In the adjoining figure, ABCD is a trapezium in which AB ∥ DC. The diagonals AC and BD intersect at O. Prove that $\dfrac{AO}{OC} = \dfrac{BO}{OD}.$

Using the above result, find the value(s) of x if OA = 3x - 19, OB = x - 4, OC = x - 3 and OD = 4.

Consider triangle AOB and COD,

∠ AOB = ∠ COD [Vertically opposite angles]

∠ OAB = ∠ OCD [Alternate angles]

So, by AA rule of similarity △AOB ~ △COD.

As triangles are similar, ratio of sides will be similar,

$\dfrac{OA}{OC} = \dfrac{OB}{OD}$

Putting values of sides from question in equation,

$\dfrac{3x - 19}{x - 3} = \dfrac{x - 4}{4} \\[1em] 4(3x - 19) = (x - 3)(x - 4) \\[1em] 12x - 76 = x^2 - 4x - 3x + 12 \\[1em] x^2 - 7x + 12 = 12x - 76 \\[1em] x^2 - 7x - 12x + 12 + 76 = 0 \\[1em] x^2 - 19x + 88 = 0 \\[1em] x^2 - 11x - 8x + 88 = 0 \\[1em] x(x - 11) - 8(x - 11) = 0 \\[1em] (x - 8)(x - 11) = 0 \\[1em] x - 8 = 0 \text{ or } x - 11 = 0 \\[1em] x = 8 \text{ or } x = 11.$

Hence, the value of x = 8 or 11.