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.