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Good evening to everybody. I was reading today a chapter on a Euclidean geometry book concerning quadrilateral constructions and there was an exercise about how to construct a quadrilateral ABCD if we are given the length of its four sides plus the length of the segment joining the midpoints of two opposite sides, that is for example if M is the midpoint of AB and N is the midpoint of DC then we know the length of MN. I tried to prove it with the extra assumption that the quadrilateral is inscribable (in that case,using Ptolemys theorem and another corollary existing in the book about the ratio of the length of the diagonals of an inscribable quadrilateral),and in this case one finds the length of both the diagonals using the data, so it is constructible.But the exercise does not say that our quadrilateral is inscribable, so I guess one cannot use the above theorems. Any ideas would be really helpful.

Petros Karajan
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  • Messing around on GeoGebra, I can see that those 5 lengths are definitely not sufficient to uniquely define a quadrilateral if you allow it to be concave... https://www.geogebra.org/classic/htr4hdrs – WW1 Jan 08 '23 at 22:06
  • Yes, of course, always we talk about concave quadrilaterals. So you think there is some mistake probably in the exercise? – Petros Karajan Jan 08 '23 at 23:03
  • read the exercise again, asking "how to construct a quadrilateral ..." is not the same as asking you to prove that it is unique. – WW1 Jan 08 '23 at 23:10
  • Ok,if we suppose it is not unique,then how can we construct it?(without geogebra of course) – Petros Karajan Jan 08 '23 at 23:38
  • Sorry,I mean convex,convex quadrilateral. Ok,problem solved, one has to bring two parallels from the midpoint M of AB,one parallel and equal to AD(say MG) and the other parallel and equal to BC(say MH). Then the triangle , lets say it MHG, is constructible,because the other midpoint N of CD is also the midpoint of HG,hence using the median theorem we can indeed construct the triangle MHG. But then again the triangles HNC and GND are constructible(the lengths of their sides are given. Connect C and D and bring parallel and equal to HC from M,we get the side AB. Hence the quadrilateral ABCD. – Petros Karajan Jan 09 '23 at 10:21

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