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 $\theta$ in the range $(0\leqq \theta <\pi /2)$, as we deform $Q$, the angle $\theta$ and the lengths of the diagonals will change and we have to prove that the area of of $Q$ is a constant multiple of $\tan \theta$. Notation: Let $\Vert x\Vert$ mean the scalar quantity of vector $x$ and the area of $Q$ be represented by $\S$. 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:

${\displaystyle S=\frac{1}{2}\Vert d_1\Vert \times \Vert d_2\Vert \sin \theta .}$

From the definition of the scalar product

${\displaystyle \Vert d_1\Vert \times \Vert d_2\Vert =\frac{d_1·d_2}{\cos \theta }.}$

So

${\displaystyle S=\frac{1}{2}\frac{d_1·d_2\sin \theta }{\cos \theta }=\frac{1}{2}{d}_1·d_2\tan \theta .}$

As shown in the October 2002 problem Flexi Quads, the scalar product of the diagonals is constant i.e.

${\displaystyle 2d_1·d_2=a_2^2+a_4^2-a_1^2-a_3^2.}$

As $a_1,a_2,a_3,a_4$, the lengths of the sides of the quadrilateral, all remain constant, hence $d_1·d_2$ remains constant. Hence the area of the quadrilateral $Q$ is a constant multiple of $\tan \theta$ and so it is proportional to $\tan \theta$.