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 θ in the range (0 ≤ θ < π/2), as we deform Q, the angle θ and the lengths of the diagonals will change and we have to prove that the area of of Q is a constant multiple of tanθ. Notation: Let |x| mean the scalar quantity of vector x and the area of Q be represented by §. In the October 2002 problem Flexi Quad Areas 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|sinθ.

From the definition of the scalar product

|d1| ×|d2| =  d1·d2 cosθ .

So

S = 1/2  d1·d2sinθ cosθ = 1/2d1·d2tanθ.

As shown in the October 2002 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 tanθ and so it is proportional to tanθ.