Proof with euclidean geometry (tangents lines) Tangents to a circumference of center O, drawn by an outer point C, touch the circle at points A and B. Let S be any point on the circle. The lines SA, SB and SC cut the diameter perpendicular to OS at points A ', B' and C ', respectively. Prove that C 'is the midpoint of A'B'.

I saw a solution by Projective Geometry. I want to know if there is a solution by euclidean geometry. I think that is possible to do with Menelaus Theorem, but I don't know wich triangles I have to use. Thanks for attention.
 A: If $\angle SAB = \alpha$ then $\angle SDB = \alpha$, $\angle OSB = 90- \alpha$ and $\angle SB'C' = \alpha$, which means $$\Delta SB'C' \sim \Delta SDB\implies {B'C'\over DB}={SC'\over SB}$$
Similary we have $$\Delta SA'C' \sim \Delta SDA\implies {A'C'\over DA}={SC'\over SA}$$
and so $$\boxed{A'C' = B'C'\iff {AD\over AS} = {BD\over BS}}$$

Since $\angle DAC = \angle DSA$ (tangent-chord) we have also
$$\Delta SAC \sim \Delta ADC\implies {AS\over AD}={SC\over AC}$$
Similary we have $$\Delta SBC \sim \Delta BDC\implies {BS\over BD}={SC\over BC}$$
and since $CA = CB$ the statement is proved.
A: Projective solution
Let $SC$ cut $AB$ and circle in $F$ and $D$. Let $AB$ meet tangents at $D$ and $S$ in $E$. Since $C$ lies on $SD$ which is polar for $E$, we see that $E$ lies on polar for $C$ which is $AB$. So $AB$ goes also through $E$, which means that $E$ and $F$ are harmonic conjugate, i.e. $(A,B;E,F)=-1$. Notice that tangent at $S$ and $A'B'$ are parallel.
Now we have:
\begin{align} 
(A',B';C',\infty) &= (SA',SB';SC',S\infty)\\
&= (SA,SB;SD,SS)\\
&=(A,B;D,S)\\
&=(DA,DB;DD,DS)\\
&=(DA,DB;DE,DF)\\
&=(A,B;E,F) \\&=-1
\end{align}
and thus $C'$ halves segment $A'B'$.

