Proving $CF = r(\frac{1-\cos x}{\sin x})$ in a geometry problem without using $\tan\frac{x}{2}$ 
This problem is from IMO $1985$ Problem $1$.

A circle has center on the side $AB$ of the cyclic quadrilateral $ABCD$. The other three sides are tangent to the circle. Prove that $AD + BC = AB$.

Let $\angle DAO$ be $\theta$. I want to show that $CG = r(\frac{1-\cos \theta}{\sin \theta})$. It is given that $$\sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \frac{1 - \cos \theta}{\sin \theta}$$ for $0 < x < 180^\circ$.
How do I use this fact to prove that $CG = r(\frac{1-\cos \theta}{\sin \theta})$? (without using the fact that $\frac{1-\cos \theta}{\sin \theta} = \tan \frac{\theta}{2}$.)
My observation:

*

*$\angle FOG = \theta$

*I also tried reverse engineering:
\begin{align}
CG = r\sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \\
CG^2 = r^2\frac{1 - \cos \theta}{1 + \cos \theta} \\
CG^2 + CG^2\cos \theta = r^2 - r^2\cos \theta
\end{align}
but I can't find why this is true.


Solution without trigonometry.
Let $P\in AB$ such that $AP=AD$.
We can assume also that $P$ is placed between $A$ and $O$ as on your picture.
Thus, $$\measuredangle APD=\frac{1}{2}(180^{\circ}-\measuredangle A)=x,$$ which says that $DPOC$ is cyclic, which gives:
$$\measuredangle CPO=\measuredangle CDO=y$$ and since $$\measuredangle B=180^{\circ}-2y,$$ we obtain $$BC=PB$$ and $$AD+BC=AP+PB=AB.$$
