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Suppose I have the figure in the image marked 'Original':

quadrilaterals divided into two triangles

Visually, that figure appears to be a parallelogram.

Would the following proof that $\bigtriangleup BCA \cong \bigtriangleup DAC$ be valid?

  1. $BC \parallel AD$, because Diagram
  2. $\angle BAC \cong \angle DCA$, because Alternate Interior Angles
  3. $AC \cong AC$, because Reflexive Property of Congruence
  4. $AB \cong CD$, because Diagram
  5. $\bigtriangleup BCA \cong \bigtriangleup DAC$, because Side-Angle-Side congruence postulate

I ask because the figure marked 'Alternate' has the same markings ($AB \cong CD$, $BC \parallel AD$), but side CD is in a different position and side AD is longer, so the two triangles are not congruent. So I'm not sure whether you can say that the original figure is in fact a parallelogram, just based on the information shown, which must be true for step 2 in the proof to be valid.

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  • $\begingroup$ Thank you. It wouldn't let me post the image as part of the question, since this is my first question on this site. $\endgroup$
    – Phil
    May 5 '17 at 18:52
  • $\begingroup$ Note that the result is still true in your 'alternate' diagram, even though multiple steps in the proof aren't, which is a good hint that the proof you're trying to give isn't the best one. $\endgroup$
    – Steven Stadnicki
    May 5 '17 at 19:01
  • $\begingroup$ So it looks like the consensus here is that this proof doesn't work with the original image. I think I'll move the parallel marks to sides AB and CD, and have step 1 refer to those sides instead. (This is a case where I'm writing/reviewing a question, rather than just trying to solve it.) $\endgroup$
    – Phil
    May 5 '17 at 19:39
  • $\begingroup$ @Phil I think the question you have in mind is whether a quadrilateral with two parallel sides and two opposite equal sides (not necessarily the same as the parallel ones) must be a parallelogram. The answer is negative, precisely because the proposed proof doesn't work. $\endgroup$
    – dxiv
    May 5 '17 at 20:40
  • $\begingroup$ @dxiv That is pretty close. It would be a bit more accurate to say the question is, "Is it reasonable / allowable to proceed with a proof in this case that treats such an image (specifically, the one shown) as a parallelogram?", but it looks like the answer to that is also negative. $\endgroup$
    – Phil
    May 5 '17 at 21:02
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Your step (5) says it's using side-angle-side, but the pictures show the configuration is side-side-angle, which does not imply congruence precisely because of this kind of counterxample.

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