Discover the fallacy in the following “proof”:

If two opposite sides of a quadrilateral

are congruent, then the remaining two sides must be parallel.

Proof

: In quadrilateral 2ABCD, let AD = BC. Assume that AB and DC are not

parallel. Let P and Q be the midpoints of DC and AB,respectively, and construct the

perpendicular bisectors at P and Q. Then these two bisectors meet in a point O. Let

{

N} =PO ∩ AB.

Since

O lies on the perpendicular bisector of CD, CO ~= DO. Similarly, OA ~= OB.

We were given that AD ~= BC, so triangleADO ~= triangleBCO by SSS congruence and angleAOD ~=angleBOC.

We can easily establish that angle

DOP ~= angleCOP. By addition, angleAOP ~= angleBOP. The

supplements of these angles must also be congruent: angleAON ~= angleBON. But, because

triangleAOQ ~= triangleBOQ by SSS congruence, angleAOQ ~= angleBOQ. Because the angle bisector

is unique, then ON and OQ must coincide and the perpendiculars to these must also

be parallel. Hence AB llCD.