Consider a quadranlge q = (ABCD) and the intersection-point E of its diagonals. Fix also a square s = (NOPQ). Draw the diagonals and the parallelograms (AEGD), (DFCE), ... created by taking the symmetric of E with respect to the middles of the sides of q. Define the affinity F mapping the created parallelogram p = (FGHI) to the square s. F maps q to some [orthodiagonal] quadrilateral q* = (JKLM) inscribed in s. The diaglonals of q* are equal and parallel to the sides of the square. Non isometric inscribed orthodiagonal quadrilaterals, like q*, cannot be immages of affinely related quadrilaterals. Indeed, going inversely from the square back and forth again to the square, by composing the relevant affinities, would give an affinity fixing the vertices of the square and interchanging the two quadrilaterals representing the affinely related quadrilaterals. This is impossible, since the only affinity fixing the vertices of the square is the identity.

This implies that every class of affinely related quadrilaterals determines a point T inside the square. Also q* is a canonical representantive of the class. Taking into account the symmetries of the square, we see that the moduli space of the classes of affinely related convex quadrilaterals is the triangle RSO.

In particular, R represents all parallelograms (they are all affinely equivalent). Segment RS represents the quadrangles which have a diagonal bisected by the other (XY is then parallel to the bisected diagonal). Finally segment RO represents all classes of trapezia.

The calculation shows that T is determined by the ratio of the segments defined on the diagonals by their intersection point E.

Look at Orthogonal_Diagonals.html for some general facts on orthodiagonal quadrilaterals.

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