[alogo] 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.

[0_0] [0_1] [0_2]
[1_0] [1_1] [1_2]

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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