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© Copyright 2002, Jim Loy
Question: Start with any general quadrilateral,
and connect the midpoints of consecutive sides, making an inscribed
quadrilateral as in the diagram. That inscribed quadrilateral, in the diagram,
seems to be a parallelogram. Let me conjecture that this inscribed
quadrilateral is a parallelogram with half the area of the original
quadrilateral. Can you prove or disprove either part of my conjecture?
Answer: Yes, the inscribed quadrilateral is always a
parallelogram, and the area is 1/2 the area of the original quadrilateral. Both
of these are well known (this is called Varignon's theorem), and so my
conjecture is not original. On the right is a diagram which answers both parts
of the conjecture.
First of all, consider the diagonals of the original quadrilateral. Each diagonal is parallel to two of the sides of our inscribed quadrilateral. This is a theorem of Euclids: In a triangle, the line connecting the midpoints of two sides is parallel to the third side. That can be easily proved by proportions and similar triangles (or other ways).
And it is also easy to prove that the parallelogram has half the area of the original quadrilateral. The two diagonals of the original quadrilateral divide it into four triangles. Let's look at the upper left triangle (in our second diagram). Our inscribed parallelogram makes a smaller parallelogram (green in the diagram) within this triangle. Draw one of the diagonals of this smaller parallelogram, the diagonal that I drew above. The parallelogram and this diagonal divide this triangle into four congruent triangles (easy to show). And so, the smaller parallelogram is half the area of this upper left triangle. The same is true for the upper right triangle and its small rectangle, and for the other two triangles and their rectangles. Summing up all of the pieces, the large rectangle has half the area of the original quadrilateral.