Prove that $\frac{AQ* RB}{QR}$ is constant, while point $P$ moves along the ray $AB$. Let $AB$ and $FD$ be chords in circle, which does not intersect and $P$ point on arc $AB$ which does not contain chord $FD$. Lines $PF$ and $PD$ intersect chord $AB$ in $Q$ and $R$. Prove that $\frac{AQ* RB}{QR}$ is constant, while point $P$ moves along the arc $AB$.
I have only a projective solution to this problem and no sythetic (and natural!) solution. Any idea? Please don't just draw some circle which will do the job with no logical explanation what was your motive to draw it. 
 A: 
Let $|AT|=a$, $|QT|=b$, $|QB|=c$,
$|AP|=u$, $|BP|=v$,
$|TP|=p$, $|QP|=q$,
$|OA|=|OB|=|OP|=|OD|=R$ (circumradius of 
$\triangle ABP$,
$\triangle ADP$, 
$\triangle DFP$, 
$\triangle FBP$).
Prove that 
\begin{align} 
\frac{(a+b)(b+c)}{b}
&=\mathrm{const},
\tag{1}\label{1}
\quad\text{ while point $P$ moves along the arc $AB$}
.
\end{align}  
\begin{align} 
\frac{(a+b)(b+c)}{b}
&=a+b+c+\frac{a\,c}{b}=|AB|+\frac{a\,c}{b}
,\\
\end{align}
\begin{align} 
\text{hence, \eqref{1} is equivalent to}\quad
\frac{a\,c}{b}&=\mathrm{const}
\tag{2}\label{2}
.\\
\triangle ATP:\quad
\frac{a}{\sin\phi}
&=
\frac{p}{\sin\alpha}
=\frac{u}{\sin(\alpha+\phi)}=\frac{2R\sin\beta}{\sin(\alpha+\phi)}
,\\
\triangle TQP:\quad
\frac{b}{\sin{\psi}}
&=
\frac{q}{\sin(\alpha+\phi)}
=\frac{p}{\sin(\beta+\theta)}
,\\
\triangle QBP:\quad
\frac{c}{\sin{\theta}}
&=
\frac{q}{\sin\beta}
=
\frac{v}{\sin(\beta+\theta)}=\frac{2R\sin\alpha}{\sin(\beta+\theta)}
,
\end{align}
\begin{align}
\frac{a^2}{\sin^2\phi}
&=
\frac{p\,2R\sin\beta}{\sin\alpha\sin(\alpha+\phi)}
,\qquad
\frac{c^2}{\sin^2\theta}
=
\frac{q2R\sin\alpha}{\sin\beta\sin(\beta+\theta)}
,\\
\frac{b^2}{\sin^2\psi}
&=
\frac{p\,q}{\sin(\alpha+\phi)\sin(\beta+\theta)}
,\\
\frac{a^2c^2}{b^2}\cdot
\frac{\sin^2\psi}{\sin^2\phi\sin^2\theta}
&=4R^2
,\\
\frac{a\,c}b&=2\,R\cdot\frac{\sin\phi\sin\theta}{\sin\psi}
=2\,R\cdot\frac{|AD|\cdot|BF|\,2\,R}{4\,R^2\,|DF|}
=\frac{|AD|\cdot|BF|}{|DF|}
=\mathrm{const}
.
\end{align}
A: Is is well known (http://www.maths.gla.ac.uk/wws/cabripages/chasles/chasles.html) that: 
1) (Chasles' Theorem)  Consider 4 fixed points $A,B,F,D$ (I follow your notations) on a circle (and more generally on a conic). For any point $P$ on the circle, the cross-ratio of the pencil of lines : 
$$[PA,PF;PB,PD] \ \text{is constant.}$$
2) The cross-ratio of a pencil of lines is equal to the cross-ratio of  intersection points with any secant transversal to the pencil, like here
$$\text{constant} = [A,R;Q,B]=(QA/QR)/(BA/BR)=(QA.BR)/(QR.BA)$$
As $BA$ is constant, $(QA.BR)/QR$ is itself constant.
$Remark :$ Is it projective geometry proof ? Although the used tools find their whole dimension in a projective context, we haven't used any typical argument of projective geometry (working up to a projective transform, sending a point to infinity, etc...)
