Well done Shu Cao of the Oxford High School for girls for producing this nice solution so promptly.
Consider any convex quadrilateral Q made from four rigid rods with flexible joints at the vertices so that the shape of Q can be changed while keeping the lengths of the sides constant. If the diagonals of the quadrilateral cross at an angle q in the range (0 £ q < p/2), as we deform Q, the angle q and the lengths of the diagonals will change and we have to prove that the area of of Q is a constant multiple of tanq.

Notation: Let |x| mean the scalar quantity of vector x and the area of Q be represented by §.

In the problem Diagonals for Area it was shown that the area of a quadrilateral is given by half the product of the lengths of the diagonals multiplied by the sine of the angle between the diagonals:
S = 1/2|d1| ×|d2|sinq.
From the definition of the scalar product
| d1| ×|d2| = d1· d2
cosq
 .
So
S = 1/2 d1· d2sinq
cosq
 = 1/2d1·d2tan q.
As shown in the problem Flexi Quads, the scalar product of the diagonals is constant i.e.
2d1 ·d2 = a22+a42-a12- a32.
As a1, a2, a3, a4, the lengths of the sides of the quadrilateral, all remain constant, hence d1 ·d2 remains constant. Hence the area of the quadrilateral Q is a constant multiple of tanq and so it is proportional to tanq.