Cyclic_Maximizing

Cyclic quadrangles have maximum area

Problem: Given the lengths a, b, c, d, to show that the cyclic quadrangle with these side-lengths has greater area than every other quadrangle with these side lengths in the same succession.

Proof based on an idea of A. Varverakis. Construct the cyclic ABCD with the given lengths. Then construct on AB a similar quadrangle ABEF to ABCD with the vertices in A, B interchanged. The result is a trapezium DCEF. Now construct some other quadrangle with the same sides in the same succession DCHG. Repeat the previous construction of GHIJ similar to DCHG and the angles at G, H interchanged. The following facts can be proved easily:
1) CD, EF and JI are always parallel,
2) triangles DGJ and CHI have equal areas,
3) the projection of I (and J) on the line carrying a is a constant point, independent of the angles x and y.
Last statement is proved by calculating the inner product of the corresponding vectors <a, b + d' >, which evaluates to a*b*cos(x) - c*d*cos(y). Later is equal to (1/2)(a^2+b^2-c^2-d^2). Later is shown in the document BasicQuadrangleIdentity.html .
4) ABDF and GHIJ have corresponding equal sides and area(DCHG)+area(HGJI) = (1+(c/a)^2)area(DCHG).
Analogously area(DCBA)+area(ABEF) = (1+(c/a)^2)area(DCBA).
5) Always area(DCHG)+area(HGJI) = area(DCIJ) <= area(DCBA)+area(ABEF).

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