# 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:

1. $$\angle FOG = \theta$$
2. 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 $$AD+BC=AP+PB=AB.$$

• +1, I did not focus on geometric solution as the question ask was very specific. Please see my edit for another variation :) Sep 20, 2020 at 17:17

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, $$AB = BC + AD$$

• I want to know how to use the fact that $\sqrt{\frac{1 - \cos x}{1 + \cos x}} = \frac{1 - \cos x}{\sin x}$, because this is provided as a hint in our exercise. but it is a nice proof why $\frac{1 - \cos x}{\sin x} = \tan \frac{x}{2}$ Sep 20, 2020 at 6:22
• Yes of course, but I suspect $\sqrt{\frac{1 - \cos x}{1 + \cos x}}$ has some geometric meaning here. Sep 20, 2020 at 6:34
• ok understood why you want to use this fact. May be that hint is just to take you to $\tan x / 2$ as $cos x = cos^2 {x/2} - sin^2 {x/2}$? Sep 20, 2020 at 6:37