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/
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