Here is the argument. Given the obtuse angle x, we make a quadrilateral ABCD with ∠DAB = x, and ∠ABC = 90◦, and AD = BC. Say the perpendicular bisector to DC meets the perpendicular bisector to AB at P. Then PA = PB and PC = PD. So the triangles PAD and PBC have equal sides and are congruent. Thus ∠PAD = ∠PBC. But PAB is isosceles, hence ∠PAB = ∠PBA. Subtracting, gives x = ∠PAD−∠PAB = ∠PBC −∠PBA = 90◦. This is a preposterous conclusion – just where is the mistake in the "proof" and why does the argument break down there?