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. ]