If the diagonals of a convex quadrilateral are divided into parts $w, x$ and $y, z$ by their point of intersection, then it is a tangential quadrilateral if and only if the angle $\theta$ between the diagonals satisfies $$\cos \theta=\dfrac{(w-x)(y-z)(2(wx+yz)-\sqrt{(w-x)^2(y-z)^2+4(wx-yz)^2})}{(w+x)^2(y+z)^2-16wxyz}$$
So please help me how to prove the statement above!
Here is what I know:
A convex quadrilateral is tangential if and only if its consecutive sides $a, b, c, d$ satisfies Pitot’s theorem, which is $a+c=b+d$.
$a^2=w^2+y^2-2wy\cos \theta$
- $b^2=x^2+y^2-2xy\cos \theta$
- $c^2=x^2+z^2-2xz\cos \theta$
- $d^2=w^2+z^2-2wz\cos \theta$