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Centroid of a Quadrilateral Construction
Posted:
Apr 5, 2006 3:30 PM
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I have found (in a text) the following curious construction of the center of gravity for an arbitrary quadrilateral ABCD.
Trisect all four sides of the quadrilateral. Name the points (clockwise, starting from A) as follows: A, J, K, B, L, M, C, N, O, D, P, Q.
Draw lines KL, MN, OP, and QJ. It is not too hard to see that they intersect in the vertices of a parallelogram.
Now the claim is that this new parallelogram has the same centroid as the original quadrilateral ABCD. Therefore, the centroid can be located by drawing the diagonals of the new parallelogram and seeing where they intersect.
The text in which I found this does not provide a proof. I have "verified" the procedure (empirically) and it looks sound, but I can't see any clear argument for why it works. Does anybody know?
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