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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?

Find the measure of the interior angles of the following regular polygons: a triangle, a quadrilateral, a pentagon, an octagon, a decagon, a 30-gon, a 50-gon, and a 100-gon.

If cos(2x) = tan^2(y), show that cos(2y) = tan^2(x).