A fairly well-known result at high school is Varignon's theorem, namely:
"If the midpoints E, F, G and H of the adjacent sides of any quadrilateral ABCD are consecutively connected, then EFGH is a parallelogram".
Perhaps less well-known is the result that the area of the Varignon parallelogram EFGH is exactly half of the area of the original quadrilateral, even in the case of the crossed quadrilateral.
Conversely, it is easy to see that if we start at the midpoint of one of the sides of ABCD and draw a line parallel to one of the diagonals it will intersect at another midpoint of an adjacent side, and continuing in this fashion the Varignon parallelogram EFGH is again obtained.
But what happens if instead we start with an arbitrary point E on AB and we continue to draw lines parallel to the diagonals as above? Do we still get a parallelogram EFGH? If so, is its area still in a constant ratio to that of the original quadrilateral, irrespective of the shape of the quadrilateral?
1) In the quadrilateral below, click on the 'Sequence 3 Actions' button. What do you notice?
2a) Drag any of A, B, C or D. Ensure that you also drag the quadrilateral into a concave as well as crossed case. What do you notice?
2b) What type of quadrilateral is EFGH? What do you notice about the ratio of its area in relation to that of ABCD (if point E remains fixed on AB)?
3) Drag E to a new position on AB and repeat Steps 1) and 2) above.
A generalization of Varignon's Theorem
4) Can you explain why (prove that) EFGH is a parallelogram?
5) Can you explain (prove) why the area of EFGH is in a fixed ratio to that of ABCD (if point E remains at a fixed position on AB)?
6) Check whether your proofs in 4) and 5) are sufficiently general to cover the concave as well as the crossed cases.
Explore More 1
7) For which position of E is the (area EFGH)/(area ABCD) ratio a maximum?
8) Challenge: Can you explain why (prove that) your observed conjecture in 7) is true for all cases (including concave & convex)?
What happens if we start with an arbitrary point E on side PQ of △PQR and we continue to draw parallel lines to the sides? Will it eventually form a closed figure, and if so, what figure would we get? If we get a closed figure, will its area still be in a constant ratio to that of the original triangle (if point A remains at a fixed position on PQ)?
9) Click on the 'Link to Triangle' button above to navigate to a new sketch showing this construction.
10) In this new sketch, click on the 'Sequence 5 Actions' button.
11) Drag any of P, Q or R. What do you notice? What type of polygon is ABCDEF? What do you notice about the ratio of its area in relation to that of △PQR (if point A remains fixed on PQ)?
12) Drag A to a new position on PQ and repeat Step 11) above.
Important: note that the area of the crossed hexagon ABCDEF is clearly NOT equal to the total area of the four triangles shaded in 'green', which is simply the representation given by the dynamic geometry software Sketchpad for the crossed hexagon ABCDEF. Similar representation issues for a crossed octagon are discussed in A Geometric Curiousity.
13) Can you explain why (prove that) ABCDEF is a closed hexagon (with opposite sides (AB & DE, BC & EF and AF & DC) parallel)?
(Remark: I first learnt of the surprising result in 13) from a talk by Shmuel Avital (Technion, Israel) in the late 70's or early 80's. This result is also mentioned in this paper of his, jointly with Ed Barbeau (Canada), in the journal for the learning of mathematics, Vol 3, Num 3, 1991, p. 5: Intuitively Misconceived Solutions to Problems. I also recently found the result mentioned in Coxeter (1961), where he mentions that apparently this crossed hexagon is called Thomson's figure in Mind (1953). However, determining its area in terms of signed areas as done here, appears to be new, or at least relatively unknown.)
14) Can you explain (prove) why the area of the crossed hexagon ABCDEF is in a fixed ratio to that of △PQR (if point A remains at a fixed position on PQ)?
De Villiers, M. (1998). A Sketchpad discovery involving triangles and quadrilaterals. KZN Mathematics Journal, 3(1), 18-21.
Some critical follow-up comments
23) As explained in De Villiers (1998), the occasional discrepancy between what Sketchpad displays regarding the area of ABCDEF and the underlying algebra determining the area, is a consequence of the software - though also using signed areas, it is taking the absolute value of the resultant area of the crossed hexagon. The reason for this is obviously to avoid negative areas. However, in hindsight it would seem better for consistency and correspondence with the underlying algebra, to not take the absolute value of the resultant area.
24) Note that technically, using signed areas correctly, the area of ABCDEF as labelled clockwise in the sketch above, should be negative (though assuming p > 0, since its area is a quadratic function of p, this area would nevertheless still become positive when A moves past the 'zero points' as shown in the sketch). Of course, the reverse would occur if we considered the area of the crossed hexagon AFEDCB.
In the 1st investigation, however, using signed areas, the area of the parallelogram EFGH would be positive since it is labelled counter-clockwise.
25) Readers are invited to repeat the above explorations on their own using other dynamic geometry software such as GeoGebra, Geometry Expressions and Cinderella. While I personally have not yet had time to repeat this investigation with these software I suspect that only Cinderella (but possibly also Geometry Expressions) would correctly & consistently use signed areas.
26) An obvious additional property of Thomson's crossed hexagon ABCDEF above is that its vertices lie on a conic (click on the 'Show Conic' button on the bottom right of the graph sketch above). This follows immediately from the converse of Pascal's theorem. For example, since the opposite sides are parallel, they pairwise intersect at points at infinity, but since all points at infinity are collinear on the line of infinity; the result follows (by the converse of Pascal).
27) The midlines (defined as the lines connecting the midpoints of opposite sides) of hexagon ABCDEF are concurrent at the centre of the conic in 26) - in this case, the centre is the centroid of PQR. For more details see Midlines Parallel-Hexagon Concurrency theorem.
28) As shown by the theorem in Perpendicular Bisectors of Parallel-Hexagon theorem the perpendicular bisectors of the sides of a parallel-hexagon (which is a hexagon with opposite sides parallel) form a parallelo-hexagon (a hexagon with opposite sides parallel and equal). This "perpendicular bisectors" theorem also directly applies to Thomson's hexagon above, but in this case, the parallelo-hexagon degenerates into a parallelogram (and is left for the reader to check).
29) Though it is not generally true that the opposite angles of all types of parallel-hexagons are always equal as mentioned in Exploration 1 at Easy Hexagon Explorations, it is valid for Thomson's crossed hexagon above (since all the angles are oriented in the same direction). Click on the 'Show Angles' button on the bottom right of the graph sketch above to view & check by dragging - the opposite angles are given in the same colors.
30) Lastly, note that all the above results for the triangle and its Thomson's hexagon (as well as for the quadrilateral and its corresponding parallelogram in the 1st investigation) generalize further if we consider moving A outside on to the extension of side PQ, and also allow the other points to correspondingly move on the extensions of the other sides. Click on the 'Link to Extended Triangle' button on the bottom right to navigate to a sketch where A has been dragged outside PQR and note that Thomson's hexagon ABCDEF then becomes convex. Drag A back on to segment PQ to see how it transforms back into a crossed hexagon.
(Note: A similar transformation from a crossed to convex hexagon occurs with the 'side-divider' hexagon discussed in Conway’s Circle Theorem as special case of Side Divider Theorem. But in the 'angle-divider' hexagon case as also discussed there, the convex hexagon transforms into a concave one if the angle divider moves outside the triangle).
Coxeter, H. S. M. (1961/1969). Introduction to Geometry. John Wiley & Sons, Inc., USA, pp. 198-199.
Coxeter, H. S. M. & Greitzer, S. L. (1967). Geometry Revisited. MAA, Washington, pp. 51-54.
De Villiers, M. (1994/2009). Generalizing Varignon's Theorem, pp. 76-88 from Some Adventures in Euclidean Geometry. Morrisville, NC: Lulu Publishers.
De Villiers, M. (1999/2012). Rethinking Proof with Sketchpad (complete PDF). Key Curriculum Press/McGraw-Hill, USA.
De Villiers, M. (2014). Slaying a Geometrical Monster: Finding the Area of a Crossed Quadrilateral. Scottish Mathematical Council Journal, 44 (Dec), pp. 71-74.
De Villiers, M. (2015). I have a dream: Crossed Quadrilaterals - a Missed Lakatosian Opportunity?. Philosophy of Mathematics Education Journal, No. 29, July.
De Villiers, M. (2020). The Value of using Signed Quantities in Geometry. Learning and Teaching Mathematics, No. 29, pp. 30-34.
Hanna, G. & Jahnke, H.N. (2002). Arguments from Physics in Mathematical Proofs: An Educational Perspective. For the Learning of Mathematics, Vol. 22, No. 3 (Nov.), pp. 38-45.
Mind, NR. (1953). Geometrical Magic. Scripta Mathematica, 19, pp. 198-200.
Oliver, P.N. (2001). Pierre Varignon and the Parallelogram Theorem. Mathematics Teacher, Vol. 94, No. 4, April, pp. 316-319.
A Geometric Paradox Explained (Another variation of an IMTS problem)
International Mathematical Talent Search (IMTS) Problem Generalized
Another parallelogram area ratio
An Area Preserving Transformation: Shearing
Some Parallelo-hexagon Area Ratios
Triangle Centroids of a Hexagon form a Parallelo-Hexagon: A generalization of Varignon's Theorem
Area Parallelogram Partition Theorem: Another Example of the Discovery Function of Proof
Finding the Area of a Crossed Quadrilateral
Crossed Quadrilateral Properties
A side trisection triangle concurrency
Conway’s Circle Theorem as special case of Side Divider Theorem
Free Download of Geometer's Sketchpad
Created by Michael de Villiers, approx. 2002, updated with WebSketchpad, 11 Oct 2023; updated 18-23 Oct 2023.