Consider a quadrilateral q = ABCD. The following remarks (exercises) lead to the proof of a identity for quadrilaterals, originally pointed out by Michael Metaxas: area( q )/tan(phi) = (1/4)(|AD|²+|BC|²-|AB|²-|DC|²).

Consider a quadrilateral q = ABCD. Draw parallels to its diagonals to build the parallologram p = FGHI. Draw the circumcircles of the resulting triangles AED, EDC, CEB and BEA. Prove the following:
1) The quadrilateral with vertices the centers of these circles p = IJKL is a parallelogram similar to p.
2) Triangles EKL, ELI, EIJ, EJK are correspondingly similar to ADC, BAD, CBA, DCB.
3) The similarity ratio of all these similar triangles (and the parallelograms) is 2sin(phi).
4) (1/2)|AC||BD|sin(phi) = area( q ) and |AC||BD|cos(phi) = |MH|² - |MG|² = (|NL|² - |NK|²)4sin²(phi) = (|JL|² - |IK|²)sin²(phi) = 2(|EL|²+|EJ|² - |EI|² - |EK|²)sin²(phi) = (1/2)(|AD|²+|BC|²-|AB|²-|DC|²).
5) area( q )/tan(phi) = (1/4)(|AD|²+|BC|²-|AB|²-|DC|²).
6) 4|AC|²|BD|² - 16area(q)² = (|AD|²+|BC|²-|AB|²-|DC|²)².
7) Conclude that a quadrilateral is [orthodiagonal] if and only if (|AD|²+|BC|²-|AB|²-|DC|²) = 0.
Look at Orthogonal_Diagonals.html for a discussion of orthodiagonal quadrilaterals.