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).

Produced with EucliDraw© |