Here's an interesting problem, and result, that I wish to share with the math community here at Math SE. I think I've found a proof without words...

enter image description here

I came across this problem sometime back, and here's a possible proof without words that I'll post as a picture -

enter image description here

I claim that one can, for any inscribed quadrilateral EFGH, make necessary constructions (several parallelograms) as in the figure, and hence show that at least one of the diagonals is parallel to a side of the quadrilateral ABCD.

Is there anything I'm missing, or is the proof complete? Let me know!

Also, please post other proofs in the answers section! (So we can all solve the problem together and discuss several methods for the same - it'll be of use to everyone to know all possible methods of approaching this problem)

I was thinking about a possible solution using complex numbers, assuming one of the side pair to be parallel to the real axis in Argand's plane, for simplicity. A similar solution, not involving too much calculation can be done using vectors! I'm not sure about coordinate geometry, as it tends to get quite cumbersome in such situations, however if anyone does manage to prove it neatly, please post the solution.

Share your ideas!

  • 1
    $\begingroup$ For your picture-proof to work, $P$, $Q$, $R$, $S$ must lie in the interior of $\square EFGH$. This isn't necessarily so: Consider $E$ and $F$ close to $A$, and $G$ and $H$ close to $C$. Granted, if the points are very close, then the inscribed area will be much less than half of the parallelogram area; it's not obvious to me that the half-area property forces the interior-points property, however. $\endgroup$ – Blue Mar 19 '18 at 14:15
  • $\begingroup$ So, what do you think? How else can I prove it? $\endgroup$ – arya_stark Mar 19 '18 at 14:16
  • $\begingroup$ Yeah, exactly. It's not obvious from the diagram, that's why I was a bit unsure about whether or not it works.. $\endgroup$ – arya_stark Mar 19 '18 at 14:17
  • $\begingroup$ It's clear by inspection that the parallelogram minus $PQRS$ is double the quadrilateral minus $PQRS$. But then you have to reflect that two quantities lose their 2:1 ratio if you add the same quantity to both of them, and this seems to require thought/words. So the picture alone doesn't seem as persuasive as e.g. some of the visual proofs of the Pythagorean Theorem. And yes, it gets worse in cases where $PQRS$ lies partly within and partly without $EFGH$. $\endgroup$ – Edward Porcella Mar 22 '18 at 18:36

I think the figure proof is applying the method of contradiction.

Suppose that 2[quad EFGH] = [//gm ABCD] and no one of the diagonals is parallel to the sides of //gm ABCD. Then, we can draw parallel lines EPS, PQH … etc to form several parallelograms as shown.

In that case, $[\triangle AFE] = [\triangle SFE] = \dfrac 12$ [//gm AFSE] and ….

Clearly, there is no region to pair up with parallelogram PQRS to make the supposition valid. If that given condition must be maintained, PQRS must be deformed to a region with zero area. Then, we have

Case-1 PQRS is made up of EP(Q)S(R)G, which is in fact EG and is parallel to the side AB.

Case-2 PQRS is made up of FS(P)Q(R)H, …..

Case-2 PQRS is a just the point T with ETG // AB and FTH // BC..


inscribed quadrialatera Here is another proof. Let parallelogram $ABCD$ be double quadrilateral $EFGH$. And suppose diagonal $EG$ is not parallel to $DC$, and diagonal $FH$ is not parallel to $AD$.

Through $F$ draw $FJ$ parallel to $AD$, and join $EJ$ and $GJ$.

Since $AFJD$ is double triangle $FJE$, and $BFJC$ is double triangle $FJG$, then $ABCD$ is double $EFGJ$. But $ABCD$ is also double $EFGH$. Therefore,$$EFGJ=EFGH$$Subtracting triangle $EFG$ from both$$\triangle EHG=\triangle EJG$$making $DC$ parallel to $EG$, which is contrary to the supposition.

Therefore, if a parallelogram is double an inscribed quadrilateral, at least one of the diagonals of the quadrilateral is parallel to a side of the parallelogram.


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