What steps need to be taken so that quadrilateral ABCD can be transformed to overlap quadrilateral A’B’C’D’ so that their vertices match?

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?

Look at attachment. has all info needed. need

Which is the graph of f(x) = (4)x? image for option 1 image for option 2 image for option 3 image for option 4