Euclidean Geometry problem: prove that $C'$ is the midpoint of $A'B'$. 
The tangents to a circumference centered at $O$, passing through an exterior point $C$, meet the circumference at the points $A$ and $B$. Let $S$ be an arbitrary point on the circumference. The lines $\stackrel{\longleftrightarrow}{SA}$, $\stackrel{\longleftrightarrow}{SB}$ and $\stackrel{\longleftrightarrow}{SC}$ intersect the diameter perpendicular to $\stackrel{\longleftrightarrow}{OS}$ at the points $A'$, $B'$ and $C'$, respectively. Prove that $C'$ is the midpoint of $A'B'$.

I'm aware that there is a relatively straightforward proof of this using Projective Geometry, but I was encouraged to solve it using Euclidean Geometry only.
I've tried to find some congruent triangles on the figure (since we're trying to prove that $\frac{A'C}{C'B} = 1$), although I couldn't find anything that would help.
I also tried the same trick used in that Menelaus' theorem proof (drop perpendiculars), which actually provided some similar triangles but was of little help too. Another idea that I had was to prove that the line parallel to $\stackrel{\longleftrightarrow}{SB'}$ passing through $A'$ and the line parallel to $\stackrel{\longleftrightarrow}{SA'}$ passing through $B'$ would meet at a point on the line $\stackrel{\longleftrightarrow}{SC}$. If I could prove that then we would be done, but then again I could not prove that either...
I would appreciate some hints/solutions to this problem.

 A: 
For convenience, take the radius of the circle to be $1$. Then,
$$\begin{align}
|OA^\prime| &= \tan \angle OSA^\prime = \tan \frac{1}{2} \angle TOA = \tan(\theta+\phi) \\
|OB^\prime| &= \tan(\theta-\phi) \\
|OC^\prime| &= \tan\angle OSC
\end{align}$$
So,
$$\begin{align}
|A^\prime C^\prime| - |B^\prime C^\prime| &= \left(\;|OA^\prime| - |OC^\prime|\;\right) - \left(\;|OB^\prime|+|OC^\prime|\;\right) \\[4pt]
&= \tan(\theta+\phi) - \tan(\theta-\phi) - 2 \tan\angle OSC \\[4pt]
&= \frac{\sin(\theta+\phi)\cos(\theta-\phi)-\cos(\theta+\phi)\sin(\theta-\phi)}{\cos(\theta+\phi)\cos(\theta-\phi)} - 2 \tan \angle OSC \\[4pt]
&= \frac{\sin 2\phi}{\cos(\theta+\phi)\cos(\theta-\phi)} - 2 \tan\angle OSC \qquad (\star)
\end{align}$$
Now, by the Law of Sines,
$$\begin{align}
\frac{|OS|}{|OC|} = \frac{\sin\angle OCS}{\sin\angle OSC} \quad&\implies\quad \frac{1}{\sec 2\theta} = \frac{\sin(\angle DOS - \angle OSC)}{\sin\angle OSC} \\
&\implies\quad \cos 2\theta = \frac{\sin 2\phi \cos\angle OSC - \cos 2\phi \sin\angle OSC}{\sin\angle OSC} \\
&\implies\quad \cos 2\theta = \frac{\sin 2\phi}{\tan\angle OSC} - \cos 2\phi \\[4pt]
&\implies\quad \tan\angle OSC = \frac{\sin 2\phi}{\cos 2\theta + \cos 2\phi} = \frac{\sin 2\phi}{2\cos(\theta+\phi)\cos(\theta-\phi)} \\
&\implies\quad (\star) = 0 \\[4pt]
&\implies\quad |A^\prime C^\prime| = |B^\prime C^\prime| \quad \square
\end{align}$$
