Carpets in a Quadrilateral II
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E, F, G, H are the midpoints of sides AB, BC, CD, DA of a convex quadrilateral ABCD. P is the intersection of AF, DE, Q the intersection of AF, BG, R the intersection of BG, CH, and S the intersection of CH, DE. Then
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Area(ΔAEP) + Area(ΔBFQ) + Area(ΔCGR) + Area(ΔDHS) = Area(PQRS).
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Explanation
Copyright © 1996-2008 Alexander Bogomolny
This problem was suggested by Bui Quang Tuan. Although simple, I do not have a recollection of seeing it anywhere else.
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The solution is a direct consequence of the Carpet's Theorem.
We split the quadrilateral in two ways
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S1  S2 = T1 T2,
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where S1 is the quadrilateral AFCH and S2 is the union of the remaining two triangles. Similarly, T1 is the quadrilateral BGDE and T2 its complement in the quadrilateral. Note that the area of each of the four regions is half that of the quadrilateral. For example,
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Area(ΔADE) = Area(ΔBDE) and
Area(ΔBCG) = Area(ΔBDG).
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Adding the two gives
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Area(ΔADE) + Area(ΔBCG) = Area(ΔBDE) + Area(ΔBDG) = Area(BGDE).
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By the Carpet's Theorem,
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Area(S1  T1) =
Area(S2 T2).
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But S1 T1 is the quadrilateral PQRS, whereas S2 T2 is the union of triangles AEP, BFQ, CGR, and DHS, as required.
The Carpets Theorem
Copyright © 1996-2008 Alexander Bogomolny
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