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Puzzle: proving a quadrilateral is a parallelogram

January 10, 2007 am31 8:48 am

Geometry teachers show their students several ways to prove that a quadrilateral is a parallelogram:

If _________________ then the quadrilateral is a parallelogram

  1. both pairs of opposite sides are congruent
  2. both pairs of opposite angles are congruent
  3. both pairs of opposite sides are parallel
  4. one pair of opposite sides is both parallel and congruent
  5. the diagonals bisect each other

Look at the following three proposed additions to the list. Some work, some don’t. For each, decide whether or not it is sufficient to prove that a quadrilateral is a parallelogram. If it is, explain why it works, if not, provide a description of a counterexample:

  1. one pair of sides is congruent, and the other pair of sides is parallel
  2. one pair of opposite angles is congruent, and one pair of sides is parallel
  3. one pair of opposite side is congruent, and one pair of opposite angles is congruent

Warning, one is very easy, one is mildly challenging, and one is fiercely hard.

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42 Comments leave one →
  1. January 10, 2007 am31 11:23 am 11:23 am

    Now you’ve piqued my interest!

    The first is easy: false. The stereotypical “gold brick” trapezoid is a counterexample (top and bottom sides horizontal and parallel; left and right are mirror images of each other).

    Two: should be true, right? I don’t remember the names of any of these rules, but let’s see: say the quadrilateral is ABCD (clockwise); AB and CD are parallel, and angles A and C are congruent. If you extend AD down through, D, the angle “below” D must be congruent to A (since AB and CD are parallel) and it is complementary (or is it supplmentary), so D is 180-A. The same logic shows that B is 180-C which is 180-A so B and D are congruent, and Q.E.D. (by your #2, I’m not sure I could actually prove that though).

    Three I believe is true, but my math is quite rusty. If you were to draw the diagonal opposite the congruent angles, you get two triangles which I believe are congruent by SSA, so the other sides must be congruent as well, yielding a parallelogram. (Is that right?? I don’t even remember which patterns prove triangle congruency, but intuitively I feel that SSA is one that does).)

    (Please forgive my horrible terminology)

  2. JBL permalink
    January 10, 2007 pm31 9:53 pm 9:53 pm

    Alas, SSA does not guarantee triangle congruence: take an isosceles triangle with angles of 120, 30 and 30, and glue an equilateral triangle to one of its legs.

    • Unknown permalink
      October 27, 2009 am31 11:15 am 11:15 am

      Can you explain this in details…..

  3. January 11, 2007 am31 12:18 am 12:18 am

    Darmok made short work of 1) and 2). The third is still open.

    (Fwiw, I didn’t think #2 was so simple. But then again, geometry’s not my thing)

  4. January 11, 2007 am31 12:32 am 12:32 am

    If you draw a diagonal connecting the non congruent angles you create triangles that are only congruent by SSA, and you don’t want to make a dummy of yourself.

    On theother hand, with #2 you have AAS, which is allowable.

  5. January 11, 2007 am31 7:05 am 7:05 am

    But can you describe (or construct) the quadrilateral with one pair of opposite angles congruent and one pair of opposite sides congruent that is not a parallelogram?

    IOW, I believe your explanation of why this should be insufficient, but I’m asking for a counterexample.

  6. January 11, 2007 am31 7:55 am 7:55 am

    Ah, JBL, thank you for that counterexample; I’m a bit embarrassed now—I should have seen that. Yes, pissedoffteacher, I suppose I did make a dummy of myself, but it’s been over a decade since I’ve done this and my skills are quite rusty.

    But I think I have it!

    But I can only describe it in a very convoluted manner, and since I don’t even have a compass anymore, I can’t be sure that my sketches are actually possible. But here goes:

    Start with a scalene triangle ABC with AB horizontal, and C nearly above B. A is 45 degrees, ABC is 100, and BCA is 35. Now place an isosceles 80-80-20 triangle to its right so that the 20 is at the top by C, and the 80s are at the base. Finally, add another copy of the original 45-35-100 triangle; the 100 angle goes by the 35+20 angle, and the 35 goes by the lone 80 of the isosceles. The 45 sticks up into the air. The two 45s are congruent, as are sides AC and its opposite.

    I have no idea if that will make sense to anyone else or if anyone has the energy to comprehend that. And if any of my friends from the hospital come over now, they are going to wonder why I have paper filled with sketches of triangles and quadrilaterals all over.

  7. January 12, 2007 am31 4:15 am 4:15 am

    I was only kidding–you are no dummy!!!!!

  8. JBL permalink
    January 12, 2007 am31 6:57 am 6:57 am

    Nicely done, Darmok!

  9. January 12, 2007 am31 11:00 am 11:00 am

    Interesting that the counterexample Darmok provided for #3 seems to involve a demonstration that SSA does not prove congruence!

  10. January 12, 2007 pm31 2:45 pm 2:45 pm

    Thanks, Pissedoffteacher and JBL!

    Actually, MRC, I based my counterexample off the SSA counterexample that JBL used to refute me above.

    Basically, what I did was start with a parallelogram, and split it into two scalenes as I described earlier. I converted the triangle on the left to its false-SSA counterpart by adding an isosceles triangle as JBL did above. The resulting triangle still had the same original angle, and those two sides remained the same length. Then I stuck the second (unchanged) triangle back on; since the diagonal was one of the Sides that was unchanged, they still aligned, and of course, the other sides and the angle were unchanged. Then I just stuck some numbers in. So I tried to refute the conjecture the same way JBL refuted me.

  11. January 12, 2007 pm31 6:25 pm 6:25 pm

    What a great problem this is! I shared it with my department and they began to realize that it was challenging. One teacher tried to construct a counterexample using GeoSketchPad but that didn’t work out. Most believed it didn’t have to be a paralleogram because of the SSA case but demonstrating an example is much harder. Here was my approach:
    I drew the diagonal connecting the non-congruent angles and proved using the Law of Sines that two of the ‘alternate interior’ angles formed had to be either equal (in which case it would be a parallelogram) or supplementary as in Darmok’s example (100 and 80!). However, I did not have Darmok’s persistence to try to draw the supplementary case by choosing a concrete example. I loved his construction. This morning I took his idea and carefully constructed a possible diagram starting with a side of 12 opposite the 100 degree angle and worked out all of the other sides using trig. I took my ruler and protractor, built it and shared it with everyone but i can’t take credit for it! Now can you name 5 of your students who would get this excited about this problem! I plan on cross-referencing all of this in my blog, MathNotations. Thanks jd2718 for challenging us!
    Dave Marain

  12. January 13, 2007 am31 2:39 am 2:39 am

    Thanks to all for attacking this one. I know it’s not a lot, but 6 math-teacherish people worked on this, and the combined and related answers that several of you provided are quite nice.

    I am struck by the richness of the discussion. Two factors:

    1) I think we owe it to ourselves and our students to put ourselves in a position of not knowing how to proceed, on a fairly regular basis. For me, this problem did exactly that. It looks like some of you felt the same way, at least initially. Too easy, and we don’t sweat, and then it is not a real problem solving situation.

    2) Several of you have blogs. It would be nice if, from time to time, you posted problems that used elementary math in novel ways, ie, easy enough to be done using high school math, but unusual or tricky enough to make a math teacher sweat. I rarely get to work on this sort of thing with interested teachers, but on one anothers blogs, maybe we might.

    Anyhow, think about it. And thanks, this was nice.

  13. January 13, 2007 am31 3:21 am 3:21 am

    Thanks for the challenge problem. I usually don’t have enough confidence in myself to answer any of these. It’s nice to read blogs and other things written by math teachers. A retired ESL teacher once said to me “wow, a math teacher that knows how to write.” I looked at her and said “An ESL teacher that knows how to tie her shoes.” She didn’t realize how insulting her comment was until I answered her back.

    Math teachers can write and do math. In fact, we can do it all!

  14. January 13, 2007 am31 3:52 am 3:52 am

    Thanks again jd…
    If you’ve had a chance to view MathNotations you know I’ve been trying to pose these kinds of questions. However, I have several other purposes which is why the daily problems I’ve posted include classroom activities which attempt to get at important ideas in mathematics as well as rich problem-solving. To add to your excellent points, perhaps the most important quality we need to develop in our students is persistence. We must encourage it, model it and offer rewards to those who display it! This quality is so rare in this time of instant information and lightning movement of fingers across cell phones, keyboards and video games. We all know Edison’s formula for genius. Seems like his message has been lost…
    Dave M

  15. January 14, 2007 pm31 1:46 pm 1:46 pm

    Thank you all and Pissedoffteacher, I don’t mind a little ribbing at being caught in what must seem like an obvious mistake to you guys. Dave Marain, any chance you’d be willing to scan your drawing? I’m very curious what the figure actually looks like; the sketches I did of it were very rough.

    Jd2718, I’m not sure if you were including me as one of the “math-teacherish people,” but I’m a physician, not a math teacher (as must be obvious by my confusion with SSA and such.

  16. January 15, 2007 am31 12:57 am 12:57 am

    I think math-teacherish can extend to mathy physicians. Why not? And yes, Dave, I have been visiting your site. You saw the 4-pairs comment, I think. I like them better when the answers are not immediately provided.

    I post a mixture of NY, teachers union, lessons, puzzles, and whatever else grabs me. But the puzzles are a constant.

    Jonathan

  17. January 15, 2007 am31 6:17 am 6:17 am

    Keep those ‘puzzles’ coming – actually it’s interesting that you’re equating challenging math problems with puzzles. I think NCTM should change the standard from problem-solving to puzzle-solving. Maybe students would prefer to work on challenging puzzles for hw!

    Darmok, I’d be happy to scan in the diagram, but I can’t insert an image into a comment. I can insert it however into my own blog or, bette yet, email it to jd or you if I had your email addresses. My email address is dmarain@rih.org if interested.

  18. January 16, 2007 am31 4:34 am 4:34 am

    Well, as long as we’re on the subject, would someone mind recapping the triangle-congruence patterns? Obviously SSA doesn’t and it’s clear that AAA wouldn’t, but I don’t remember the rest.

    Thanks, Dave, I just e-mailed you.

  19. JBL permalink
    January 20, 2007 am31 2:55 am 2:55 am

    SSS does. SAS does. SSA *almost* does (only one extra), but we’ve covered that. AAA doesn’t, but it does determine the triangle up to scaling (“similarity”), so as a result, knowing AAA and *any* corresponding length gives you congruence. As special cases of this we have SAA and ASA (since if I know AA, I know AAA because the sum of the angles is constant).

    I just ran across the following AIME problem from 2003, which reminded me of this thread:

    “ABCD is a convex quadrilateral with AB = CD = 180, perimeter 640, AD ≠ BC, and ∠A = ∠C. Find cos A.”

    (I hope that formats correctly.) It can’t be solved just with the tools we’ve discussed (one extra tool is needed, plus some algebra), but it’s clearly related in flavor.

  20. January 20, 2007 am31 3:13 am 3:13 am

    SS+acute angle = SS+acute angle works
    SS+obtuse angle = SS+obtuse angle works
    and
    SS+right angle = SS+right angle works (usually called “Hypotenuse Leg”

  21. January 20, 2007 pm31 1:25 pm 1:25 pm

    Thanks! Great refresher and the SAA and ASA explanation will help me to remember that. So there are no other possibilities, right? AAA and SSA are the only ones that don’t guarantee congruence?

  22. Jaimie 09' permalink
    January 26, 2007 am31 4:38 am 4:38 am

    Jaimie was here. yo yo yo 09′

  23. February 12, 2007 am28 2:01 am 2:01 am

    if you take two congruent triangles and rotate one of them 180 degrees, and then glue it back to the original one, you get a parallelogram. since SSA doesn’t work, find two triangles that are not congruent, but conform to SSA, and then do the same rotate-and-glue thing, and you’ll get a counter-example. the first one i tried was concave, but that should count as a counter-example too!

  24. February 12, 2007 am28 2:42 am 2:42 am

    Yup, that’s precisely the approach we used here to construct a counterexample, Polymath (see, for instance, my comment #10 above). Incidentally, my first construction was concave as well!

  25. February 20, 2007 pm28 11:29 pm 11:29 pm

    A problem which you discuss on your site is very important for me. Thank you for your resume.

  26. J.D permalink
    August 20, 2007 pm31 12:34 pm 12:34 pm

    This is actually just what my homework is about!

  27. NeedHelpWithMathDude permalink
    August 28, 2007 am31 6:13 am 6:13 am

    i have a question thats been killing me cuz im dumb and i am a little new to a few of the geometric concepts, but…..

    if i have a quadrilateral that both sets of opposite sides are congruent but not congruent with each other, can i prove its a parallelogram?

    ____|_____
    \ \
    = =
    \____|____\ the “|” represents the first set of congruent lines,
    and the “=” represents the second

  28. NeedHelpWithMathDude permalink
    August 28, 2007 am31 6:13 am 6:13 am

    crap the above message drawing did not come out correct… srry

  29. NeedHelpWithMathDude permalink
    August 28, 2007 am31 6:15 am 6:15 am

    this should be right i hope

    ____|_____
    \………….\
    =………….=
    \____|____\

  30. NeedHelpWithMathDude permalink
    August 28, 2007 am31 6:16 am 6:16 am

    well…. u get the idea :(

  31. August 28, 2007 am31 6:30 am 6:30 am

    Indeed, the idea is clear. Consider a quadrilateral ABCD with AB \cong CD and BC \cong DA (that’s what you were trying to draw).

    Draw diagonal AC. Now \triangle{ABC} \cong \triangle{CDA} by side-side-side. Do you see it?

    Now \angle{BCA} \cong \angle{DAC} (corresponding parts of congruent triangles are congruent). With me?

    Extend AD and extend BC, (this is just to help you see, not part of the proof). They are two long lines, cut by a transversal (AC) and alternate interior angles are congruent (we just proved that, above), therefore, the two lines are parallel.

    Pick the other two interior angles to prove AB || CD, and now we’ve met the definition of a parallelogram.

  32. sexybabe mwah permalink
    September 28, 2007 pm30 11:36 pm 11:36 pm

    how do u prove that the angles on a triangle add up to 180 degrees

  33. September 29, 2007 am30 5:12 am 5:12 am

    That depends on what you already know.

    The last time I showed it to somebody, we took any old triangle ABC, and drew a line parallel to BC through A (call the line DAE, with D and B to the left, E and C to the right)

    \angle{DAB} + \angle{BAC} + \angle{CAE} = 180 (they make a straight line)

    but \angle{DAB} = \angle{B} since they are alternate interior, and likewise \angle{CAE} = \angle{C}

    So those two, plus BAC, make 180, and those two, plus BAC, are the interior angles of an arbitrary triangle.

  34. December 4, 2007 pm31 10:22 pm 10:22 pm

    We do these, with the exception of the 1 pair parallel and 1 pair of congruent angles proof, every time we teach Geometry at my high school. It’s built in to the Contemporary Mathematics In Context Curriculum. http://www.wmich.edu/cpmp/

    I may add that proof this time through, so thanks for the problem!

  35. February 5, 2008 am29 4:16 am 4:16 am

    lol theres comments on tryangles

  36. Garrett permalink
    November 15, 2008 am30 5:29 am 5:29 am

    hahahahah
    what a bunch of math nerds!!!!
    yall have no life

  37. jeric permalink
    January 13, 2009 am31 8:05 am 8:05 am

    i dont agree

  38. February 11, 2010 am28 9:48 am 9:48 am

    very good information , good teaching is really the best way for training

  39. Sam permalink
    May 17, 2010 pm31 5:43 pm 5:43 pm

    I know I’m late to the party, but isn’t there another counter-example for 3?

    Here is an image:

    http://tmsam.blogspot.com/2010/05/re-httpjd2718wordpresscom20070110puzzle.html

  40. January 11, 2013 pm31 2:56 pm 2:56 pm

    “Puzzle: proving a quadrilateral is a parallelogram JD2718”
    actually got me personally addicted with your web site! I personallywill wind up being
    returning much more frequently. Many thanks -Janette

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