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### Proving Quadrilateral is a Parallelogram

```
Date: 11/30/2001 at 12:24:10
From: Kara
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 opposite 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

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/
```
Associated Topics:
High School Constructions
High School Geometry
High School Triangles and Other Polygons

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