In quadrilateral ABCD the diagonals AC and BD meet at O. Which one of these conditions, taken on its own, does NOT force ABCD to be a parallelogram?

Diagonals that bisect each other, both pairs of opposite sides equal, and one pair of opposite sides both parallel AND equal are each enough on their own — every one of the three is a standard test for a parallelogram. But one pair parallel together with the OTHER pair merely equal is not enough: an isosceles trapezium has AB ∥ DC and AD = BC yet is certainly not a parallelogram. Pairing a parallel condition with an equality about the wrong pair of sides is the classic slip.