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.

 A: 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 $$AD+BC=AP+PB=AB.$$
A: I am assuming you are trying to prove $AO = AE + CG$ and $OB = DE + BG$.
Coming to your specific question, if you do not want to use $\frac{\theta}{2}$, make a construct with $\theta$.
$OH = r \cos \theta$
$CI = CF \sin \theta = OG - OH$
As $CF = CG$,
$CG \sin \theta = r - r \cos \theta$
$CG = r\frac {1 - \cos \theta} {\sin \theta}$

EDIT: Motivated by Michael Rosenberg's solution without trigonometry but I think somewhat simpler,
Place a point $P$ on line $AD$ (extend if required) such that $AP = AO$.
Then $\, \triangle OPE \cong \triangle OCG$ (Angle-Angle-Angle, one side same)
So, $AO = AP = AE + EP = AE + CG$
Similarly, $OB = BG + DE$
Hence, $AB = BC + AD$
