Proving Quadrilateral Is a Parallelogram, ReduxDate: 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 angles 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: Proving Quadrilateral is a Parallelogram 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: http://mathforum.org/dr.math/gifs/dave04.04.12.jpg 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/ |
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