Integral Area For Sure?

Let the length of diagonals $AC, BD$ be $a, b$ respectively, and let the angle formed between the diagonals (which includes the horizontal line that passes through the intersection point of the diagonals) be $θ$ . Let the gradients of $AC, BD$ be $m, -m$ respectively.

We have $θ = 2 \tan^{-1} m$ . This is because the angle $θ$ is twice the angle formed by AC and the x-axis.

Now the area of the quadrilateral ABCD, $S = \frac{1}{2} ab \sin θ$ $= \frac{1}{2} ab \cdot \frac{2 \tan(\tan^{-1} m)}{1+\tan^2 (\tan^{-1} m)}$ $= \frac{abm}{1+m^2}$

Now let $m= \frac{p}{q}$ , then the above expression becomes $\frac{abpq}{p^2+q^2}$

Since $A, B, C, D$ are lattice points, we may partition $AC$ and $BD$ into an integral number of portions which, by Pythagoras' theorem, has length of $\sqrt{p^2+q^2}$ . Hence it is proven that the area of the quadrilateral $ABCD$ must be an integer.