Question

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?

Answer 1

Answer:

The error in the argument is in assuming that the line from p to d passes through line ab - that is, that pd crosses below a. But actually pd passes above a. This means that the angle pad is not equal to x + pab, but rather angle pad = 360-(x+pab). The rest of the "proof" then falls apart.

Step-by-step explanation: