Anyway, my idea was something like this:
Assume that the minor arc DE has “angle measure” of x^{\circ}, then can you show that \angle ECD = \angle ACB = \frac{180^{\circ} + x^{\circ}}{2} = \left(90 + \frac x2\right)^{\circ}? Since we know that x > 0 (unless we consider the degenerate case of D being coincident with E; this case is not interesting because you already know what happens here), then \left(90 + \frac x2\right)^{\circ} > 90^{\circ} implying \angle ACB > 90