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A Parallelogram?

© Copyright 2002, Jim Loy

I received email from someone who said that he had discovered a new theorem: In a quadrilateral, when two opposite sides are equal and two opposite angles are equal, then the quadrilateral is a parallelogram. He also said that he hadn't been able to prove it. And here we see why. The new theorem is not true, as I sent him a picture of the quadrilateral on the left, where angle A = angle C, and AB = CD. We see more detail on the right. We draw any obtuse triangle ABF, with obtuse angle F. We then extend AF, and draw EB perpendicular to AF, and make ED = EF. Then we make DC = AB and BC = AF. Then we can show that triangle ABF is congruent to triangle CDB (See Congruence Of Triangles, Part I). So angle A = angle C.

Given a triangle like ABF above, there are often two quadrilaterals which meet the above conditions, one is a parallelogram (with a vertex at F), and the other is not. This ambiguity is related to the side-side-angle (SSA) non-theorem for congruence of triangles, which is also ambiguous. In our diagram, the two triangles ABF and ABD have two sides and one angle equal to each other, but are not congruent.

Addendum:

I find that this false theorem is not even new. P. Halsey published this false proof in 1959, as an example of a convincing false proof.

False proof: Here angle A = angle C, and AB = DC. We draw BE perpendicular to AD at E, and DF perpendicular to BC at F. Draw BD. Triangles ABE and CDF are congruent, so BE = DF and AE = CF. Then triangle BDE is congruent to triangle DBF, so ED = FB. So AE+ED = BF+FC or AD = BC. Since AB = DC and AD = BC, the quadrilateral is a parallelogram. Can you find the flaw in the proof?

Flaw: The flaw in the proof is that E or F may be outside the quadrilateral, and so AD is not always AE + ED, and BC is not always BF + FC.

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