Carpets in a Quadrilateral II

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

Explanation

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

We split the quadrilateral in two ways

S1 S2 = T1 T2,

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,

Adding the two gives

By the Carpet's Theorem,

Area(S1 T1) = Area(S2 T2).

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
• 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