Theorem: 0 = 1 Proof: (The notation "Y =~ Z" means "Y is congruent to Z".) Construct quadrilateral ABCD as follows: (1) angle ABC is a right angle; (2) angle BCD is slightly obtuse--say, 100 degrees; (3) side AB =~ CD. (Note that AD and BC aren't quite parallel, since D is slightly closer to the line BC than A is) Construct the perpendicular bisectors of BC and AD. Since AD and BC are not parallel, their perpendicular bisectors aren't parallel, and thus meet at a point X. (This point X will be below the middle half of the drawing. The same "proof" holds whether X is inside or outside ABCD.) [Sorry, too hard to draw in ascii.] First we show that triangle ABX =~ DCX: AB =~ DC, by construction. BX =~ CX, since X is on the perpendicular bisector of BC. AX =~ DX, since X is on the perpendicular bisector of AD. Thus triangle ABX =~ DCX (since they have three corresponding congruent sides). Therefore angle ABX =~ angle DCX (corresponding angles of congruent triangles). But also, angle XBC =~ angle XCB (since they are the base angles of isosceles triangle XBC). Thus we have that right angle ABC is congruent to obtuse angle BCD. So 90 degrees = 100 degrees; it follows immediately that 0=1 as claimed. [ I heard this in a class taught by Kris Sikorski in 1985 (U of Utah, cs dept), don't know if it's original with him or not. ]