According to the journal of irreproducible results, any obtuse angle is a right angle! c d b p 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 = p d. so the triangles p ad and p bc 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?
The mistake stems from the assumption that angle dab and abc are both 90 and ad = bc and that the perpendicular bisector of dc is different from the perpendicular bisector to ab because they are the same and abcd is a rectangle.
Step-by-step explanation:
If ∡dab = ∡abc and side ab is equal to side bc which are opposite sides, ten then ab is parallel to bc which means the quadrilateral is parallelogram. Also since two angles of the four angles of the parallelogram are 90 degrees then the parallelogram is a rectangle.
The bisector of one side of a rectangle will also bisect the opposite side of the rectangle. Therefore the bisector of dc is the same as the bisector of ab and it meets ab at the midpoint of ab. Therefore p is now at the midpoint of ab and there are no triangles pad and pbc.
