2
$\begingroup$

quadrilateral *n*

Let quadrilateral n have vertices A, B, C, and D. If side AB is congruent to side CD, and angles ABC and CDA are also congruent. Can you show this is not necessarily a parallelogram?

It does not follow the conditions for a parallelogram, yet my math teacher and I were unable to show a counter example or prove that it is or is not.

One cannot dissect it and use triangle congruence, for it leads to ASS.

Can someone show it is not a parallelogram, or even better, show a counter example?

  • 0
    Are you secretly assuming it is a convex quadrilateral? Because a nonconvex one is an obvious counterexample. – 2017-02-23
  • 0
    rschwieb - there was a visual in the given situation that implied it was convex – 2017-02-24
  • 0
    typically in geometry problems visuals are taken with a grain of salt as to what they imply. – 2017-02-24
  • 0
    rschwieb I would like to see your concave counter example if you could make an image. – 2017-02-24
  • 0
    You're right about the visual, too, but this was a first year geometry test very early in the year(very cryptic imagery that could still be accurate wasn't really present) – 2017-02-24
  • 0
    Isn't it obvious what a nonconvex example would be? Something like ⋈. – 2017-02-24

1 Answers 1

3

Yes, it can be a non-parallelogram.

enter image description here

Regarding the triangles $\triangle ABC$ and $\triangle CDA$ - due to the two solutions for the inverse sine, $\measuredangle DAC$ and $\measuredangle ACB$ can be different.

Note that if the given $\measuredangle ABC = \measuredangle CDA$ is obtuse, there is only one viable solution for the inverse sine and the shape is indeed a parallelogram.

  • 0
    Thanks for the counter example. I haven't done much work with sine and cosine yet, but now I can see how the 2 triangles viable with ASS can interact. – 2017-02-24