Is a rectangle always a parallelogram?

Here is their 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?

List x1, x2, x3, x4 where xi is the midpoint endpoint of the five equal intervals used to estimate the area under the curve of f(x) between x = 0 and x = 10.

Aweekly newspaper charges $72 for each inch in length and one column width. what is the cost of an ad that runs 2 column inches long and 2 columns wide? its not 288