Home > Challenge, Geometry, Teaching > Quadrilateral Challenge — A Solution

Quadrilateral Challenge — A Solution


Here is one approach to answering the quadrilateral challenge posed earlier this week.  In summary, the challenge was to prove or disprove the following statement:  A quadrilateral with a pair of congruent opposite sides and a pair of congruent opposite angles is a parallelogram.

I offer this disproof without words.

By starting with an isosceles triangle, cutting it, rotating one of the pieces, and gluing it back together, we have constructed a quadrilateral with one pair of congruent opposite sides and one pair of congruent opposite angles that it is not necessarily a parallelogram!

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  1. December 16, 2011 at 11:24 am

    This a great exercise – because I can use it, I’m thinking, with my poetry unit. Do you have a sheet with those diagrams? Or did you just project the wordless proof on the board? Congruent angles are words, the area of triangles the form and content of the poem, and the syntax is the one variable that determines meaning… something… some parallel…

    • December 16, 2011 at 4:54 pm

      I like your idea, what I understand of it, anyway. I’ll gladly provide you a copy of the diagrams if you can further elucidate your theory of geometric poetry.

    • NA
      December 17, 2011 at 4:44 pm

      I would love to see this in class, Mr. Ree! 🙂

      • December 20, 2011 at 4:43 pm

        Thanks for the article, JBL! I’ve passed it along; it’s a quick, amusing read.

  2. JBL
    December 16, 2011 at 12:19 pm

    And, to tie this in to my comment on the previous thread, cutting an isosceles triangle by a cevian is exactly how you get a pair of noncongruent triangles that are related by SSA.

    • December 16, 2011 at 4:58 pm

      Yes. I love the appearance of the Law of Sines ambiguous case here. This also offers another explanation as to why, if the opposite congruent angles are right, you can prove it’s a parallelogram.

      • JBL
        December 17, 2011 at 7:15 pm

        Nice; I had been wondering whether there was a “good” reason that right angles had that property (and equivalently why SSA *is* a congruence relation for right triangles); tying in the Law of Sines and the ambiguity of arcsin is a nice way to make the connection.

        Somewhat further afield, this also settles the question of whether such a quadrilateral can be cyclic: if so, the opposite equal angles must be right, whence we’re stuck with just rectangles.

  3. moukaouame
    December 16, 2011 at 4:46 pm

    Il faut rajouter la convexité.

    • December 16, 2011 at 4:56 pm

      My French isn’t so good: are you suggesting that we add the requirement of convexity to the quadrilateral to eliminate this counterexample?

      • JBL
        December 17, 2011 at 12:14 am

        Yes, that’s the suggestion. But it doesn’t eliminate all counter-examples: you happened to choose an example that involves gluing two obtuse angles together, but it’s easy enough not to. For example, if you cut an equilateral triangle with a cevian that trisects one side and apply this operation, the resulting quadrilateral is convex.

  4. December 17, 2011 at 2:46 pm

    Nice example. You inspired me to cut up some equilateral triangles and explore!

    • Yelena Weinstein
      December 18, 2011 at 9:06 pm

      Yes! That’s what I was trying to draw on the board in MfA! But it’s a cool and convincing way.

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