Proving Quadrilateral is a Parallelogram

Date: 11/30/2001 at 12:24:10
From: Kara Curran
Subject: Proving Quadrilateral is a parallelogram

We are having a problem with the idea of a quadrilateral having one 
pair of opposite sides congruent and one pair of opposite angles 
congruent. We've flipped the figure and found isosceles triangles, so 
we THINK that we can say a quadrilateral is a parallelogram if these 
conditions exist. 

Is there a theorem that states you only need one pair of oppoite sides 
and angles congruent for it to be a parallelogram? 

If there isn't, can you just give us a clue to how we can disprove it?

Date: 11/30/2001 at 16:44:04
From: Doctor Rob
Subject: Re: Proving Quadrilateral is a parallelogram

Thanks for writing to Ask Dr. Math, Kara.

The statement you want to be true isn't. Here is a way to construct a 
counterexample.

Draw an acute angle <A. From some point B on one side, using a large 
enough radius, draw an arc of a circle that intersects the other side 
in two points C and D:

                         B _,-'
                       _,o'
                   _,-' / \
               _,-'    /   \
           _,-'       /     \
       _,-'          /       \
   A o--------------o---------o-----
                    D         C

By construction BC = BD, since they are radii of the same circle.
Notice that AC > AD. Now make a copy EFG of triangle ABD, and erase 
line segment BD and point D:

                         B 
                       _,o
                   _,-'   \
               _,-'        \
           _,-'             \
       _,-'                  \
   A o------------------------o C

                         F
                       _,o
                   _,-' /
               _,-'    /
           _,-'       /
       _,-'          /
   E o--------------o
                     G

Rotate it around so that F coincides with C and G coincides with B.  
This is possible because FG = BD = BC = CB. Then you have a 
quadrilateral like this:

                                E
                                o
                               /|
                              / |
                             /  |
                            /  |
                           /   |
                      B=G /    |
                       _,o     |
                   _,-'   \    |
               _,-'        \  |
           _,-'             \ |
       _,-'                  \|
   A o------------------------o
                             C=F

Now erase line segment BC, and you have a quadrilateral ACEB.

                                E
                                o
                               /|
                              / |
                             /  |
                            /  |
                           /   |
                        B /    |
                       _,o     |
                   _,-'        |
               _,-'           |
           _,-'               |
       _,-'                   |
   A o------------------------o
                               C

By construction EC = EF = AB, and <A = <E, so two opposite sides are 
equal and two opposite angles are equal, but this is definitely NOT a 
parallelogram, because EB = EG = AD < AC.

You may object that the above quadrilateral is not convex, that is,
it has an interior angle greater than 180 degrees. In this
counterexample, that is true. If <A had been chosen larger, and points 
C and D closer together, a convex quadrilateral satisfying your 
conditions, but still not a parallelogram, would have resulted.

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/