# 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

Area(ΔAEP) + Area(ΔBFQ) + Area(ΔCGR) + Area(ΔDHS) = Area(PQRS).

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Copyright © 1996-2017 Alexander BogomolnyThis problem was suggested by Bui Quang Tuan. Although simple, I do not have a recollection of seeing it anywhere else. (P.S. Years later I came across this problem in the highly recommended Mathematical Olympiad Treasures by T. Andreescu, B. Enescu (Birkhäuser, 2004).)

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The solution is a direct consequence of the Carpet's Theorem.

We split the quadrilateral in two ways

S_{1} ∪ S_{2} = T_{1} ∪ T_{2},

where S_{1} is the quadrilateral AFCH and S_{2} is the union of the remaining two triangles. Similarly, T_{1} is the quadrilateral BGDE and T_{2} its complement in the quadrilateral. Note that the area of each of the four regions is half that of the quadrilateral. For example,

Area(ΔADE) = Area(ΔBDE) and

Area(ΔBCG) = Area(ΔBDG).

Adding the two gives

Area(ΔADE) + Area(ΔBCG) = Area(ΔBDE) + Area(ΔBDG) = Area(BGDE).

By the Carpet's Theorem,

Area(S_{1} ∩ T_{1}) =
Area(S_{2} ∩ T_{2}).

But S_{1} ∩ T_{1} is the quadrilateral PQRS, whereas S_{2} ∩ T_{2} is the union of triangles AEP, BFQ, CGR, and DHS, as required.

### Carpets Theorem

- The Carpets Theorem
- Carpets in a Parallelogram
- Carpets in a Quadrilateral
- Carpets in a Quadrilateral II
- Square Root of 2 is Irrational
- Carpets Theorem With Parallelograms
- Two Rectangles in a Rectangle
- Bisection of Yin and Yang
- Carpets in Hexagon
- Round Carpets
- A Property of Semicircles
- Carpets in Triangle
- Carpets in Triangle, II

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Copyright © 1996-2017 Alexander Bogomolny61186176 |