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 \leq 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

\tan θ

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:

$S = \frac{1}{2} | d_{1} | \times | d_{2} | \sin θ .$

From the definition of the scalar product

$| d_{1} | \times | d_{2} | = \frac{d_{1} \cdot d_{2}}{\cos θ} .$

So

$S = \frac{1}{2} \frac{d_{1} \cdot d_{2} \sin θ}{\cos θ} = \frac{1}{2} d_{1} \cdot d_{2} \tan θ .$

As shown in the problem Flexi Quads, the scalar product of the diagonals is constant i.e.

$2 d_{1} \cdot d_{2} = a_{2}^{2} + a_{4}^{2} - a_{1}^{2} - a_{3}^{2} .$

As $a_{1}, a_{2}, a_{3}, a_{4}$ , the lengths of the sides of the quadrilateral, all remain constant, hence $d_{1} \cdot d_{2}$ remains constant. Hence the area of the quadrilateral $Q$ is a constant multiple of $\tan θ$ and so it is proportional to $\tan θ$ .