TITLE: Locating three sets of collinear points
QUESTION [15 upvotes]: Given any three distinct points $A,B,C$ and a circle $C(O)$, construct points $D,E,F$ on the circle such that 

$A,D,E$ are collinear,
$B,E,F$ are collinear and,
$C,F,D$ are collinear. 

One such solution is indicated on the diagram below. I have enough analytic and numerical evidence to indicate that these points exists. In fact, there are two such sets of points, as shown by @coproc below. However I would like a geometric construction. 

One idea: My idea was to invert the points $A,B,C$ in the circle to find points $A',B',C'$ respectively. The problems then becomes equivalent to the following problem. 
Given three points $A',B',C'$ and a circle $C(O)$. Construct three circles such that 

One circle passes through points $A'$ and $O$,
another passes through points $B'$ and $O$,
the third passes through points $C'$ and $O$ and,
the three circles intersect pairwise on $C(O)$. 

The required points $D,E,F$ are just the points of intersection of these three circles.

REPLY [2 votes]: [this is an addition to mercio's first answer: for my understanding there was still a proof missing, why the involved involutions on the circle like $\sigma_A$ (see mercio's answer or below for the precise definition) can be represented as a function of the form $t \mapsto \frac{at+b}{ct+d}$ in the projective line; so I want to share my results on this; I am not aware of any shorter argument]
A circle can be parametrized using stereographic projection. For our needs here it is easiest to consider the unit circle with center $M=(0,1)$:

The real parameter $t$ parametrizes the circle $x^2 + (y-1)^2 = 1$ by 
$$t \mapsto U(t) := \left(\frac{2t}{t^2+1},\frac{2}{t^2+1}\right) .$$ 
For a given point $(x,y)$ on the circle we get back $t$ by $t = \frac{x}{y}$. In this parametrization only the point $P$ is missing. The parametrization can be completed by going to homogenous coordinates $[t:1]$ and assigning $[1:0]$ to $P$.
Now let us consider the special situation when $A$ is on the $y$-axis. The involution $\sigma_A$ maps a point $T_0 = U(t_0)$ on the circle to the other intersection of the line $AT$ with the circle $T_1 = U(t_1)$. 

Now how are $t_0$ and $t_1$ related? It turns out that
$$t_1 = \frac{y_0-2}{y_0t_0} ,$$
where $y_0$ is the $y$-coordinate of $A$. (It takes some algebra to compute the second intersection of the circle $x^2 + (y-1)^2 = 1$ with the line through $A$ and $T_0$, whose equation is
$$ y = \frac{\frac{2}{t_0^2+1}-y_0}{\frac{2t_0}{t_0^2+1}}x+y_0 .)$$
So in the special case of $A$ on the $y$-axis the involution $\sigma_A$ can indeed be represented in the form $\frac{0\cdot t+(y_0-2)}{y_0t+0}=\frac{at+b}{ct+d}$.
For the general case of the position of point $A$ it suffices to check that rotating a point on the circle corresponds to a transformation of parameter $t$ of the same form.

Now here the homogenous coordinates come in handy. From the above figure it can be seen that
$$
U([t_1:h]^T) =
\left(\begin{array}{cc} \cos\alpha & -\sin\alpha \\
                        \sin\alpha &  \cos\alpha
\end{array}\right) U([t_0:h]^T)
=
U\left( 
\left(\begin{array}{cc} \cos\frac{\alpha}{2} & -\sin\frac{\alpha}{2} \\
                        \sin\frac{\alpha}{2} &  \cos\frac{\alpha}{2}
\end{array}\right)
[t_0:h]^T
\right)  
,$$
which shows that the transformation of $t_0$ to $t_1$ again is of the expected form. Since also concatentions of functions of the form  $t \mapsto \frac{at+b}{ct+d}$ are of this form the - at least for me - missing details seem to be filled.