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### Proving Quadrilateral Is a Parallelogram, Redux

```Date: 04/04/2012 at 22:06:07
From: Dave
Subject: Proving that a quadrilateral is a parallleogram

I have a student who conjectured that a quadrilateral with one pair of
congruent sides and one set of congruent angles is a parallelogram.

A Dr. Math conversation begun by Kara on 11/20/2011 delivers a
counterexample given for a non-convex case, with which I agree. But my
sophomore geometry student persisted, and has produced a proof. What I
believe to be interesting is that this problem does not appear in any
textbooks, so it could be an original proof. That would be very exciting
for my student!!

Could you see if you agree that it is valid?

Here it is.

Given a quadrilateral ABCD with <D congruent to <B and AD congruent to BC.

Drop a perpendicular segment from A to CD at E, and a perpendicular
segment from C to AB at F.

Triangle AED is congruent to triangle CFB by AAS. Thus DE is congruent to
FB; and AE is congruent to CF because corresponding parts of congruent
triangles are congruent (CPCTC).

Construct AC in quadrilateral AFCE. Right triangles ACF and CAE are
congruent by the hypotenuse leg postulate (HL). Thus AF is congruent to EC.

By angle addition and substitution, CD is congruent to AB. Then the
original quadrilateral ABCD has two pairs of congruent sides, so it must
be a parallelogram.

```

```
Date: 04/04/2012 at 23:42:20
From: Doctor Peterson
Subject: Re: Proving that a quadrilateral is a parallleogram

Hi, Dave.

Very nice -- but there's a subtle flaw in the proof.

The error is in this line:

By angle addition and substitution, CD is congruent to AB.

This is based on the facts that CE is congruent to AF, and that DE is
congruent to BF, and the UNSTATED ASSUMPTION that ...

AF = AB + BF and CE = CD + DE,

... so that ...

CD = CE - DE
= AF - BF
= AB

This is true ONLY IF B lies between A and F, and D lies between C and E.

To show that this need not be true, I followed the instructions in the
page you referred to:

http://mathforum.org/library/drmath/view/55401.html

I drew a counterexample (and arranged it, as suggested at the bottom, so
that the quadrilateral is convex, just to avoid any concerns in that
area). Here is my result; look carefully and you'll see that the
betweenness assumption is false here:

This is very hard to catch unless you are aware of the danger, because you
naturally draw the figure as you expect it to look (in this case, a
parallelogram), and without a good bit of experience you don't tend to
consider other possibilities.

I immediately guessed what might be happening because I've seen the same
thing in some classic fallacious proofs:

Are All Triangles Isosceles?
http://mathforum.org/library/drmath/view/55192.html

Every Triangle Is Isosceles!?
http://www.math.wisc.edu/~robbin/461dir/isosceles.pdf

A Right Angle Is Equal To an Obtuse Angle?
http://www.jimloy.com/geometry/obtuse.htm

It may not be as exciting as discovering a new and valid proof, but you
may have here a new and rather beautiful fallacious proof. (I can't
guarantee, however, that it hasn't been done before.)

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Constructions
High School Euclidean/Plane Geometry
High School Triangles and Other Polygons

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