Suppose $T$ is intersection of $QR$ and $AB$ If the tangent from $T$ to circle (not $AB$ the other) is' not parallel to $AC$ by Brianchon theorem $PR$ , $AC$ and tangent from $T$ to circle are concurrent. But $PR$ is parallel to $AC$ contradiction. So $T$ is constant.

Well done . A very nice solution. See my solution( its a little bit longer): I will prove that the pole of QR is on a fixed line. The pole of QR is the intersection of the polars of Q and R. Let L be tangent to C in X. So the polar of Q is XC . we name the line through P parallel to AC: d. So the polar of R is the line through A and the pole of d .The pole of d (which is Y) is on the diameter of C perpendicular to d( which is the diameter through C which we name CD) .it lies also on BX. So we have found Y=( CD, BX). Let Z be the intersection of CX and AY( so Z is the pole of QR). Now to finish: its obvious that Z is on BD which is constant ( because AY ,the polar of R, passes through Z so the polar of Z passes through R so Z is on BD).

Here's what I did: we must show that the application from line $AC$ to line $BC$ which maps $Q\to R$ is a perspectivity. In order to do this we must show that it's a homographic (preserving the cross-ratio) transformation and it maps $C\to C$. It's easy to see that it's homographic because it's the composition of $Q\to P$ (from line $AC$ to line $AB$; it's wel-known that this sort of transformation is homographic (even if $(C)$ is a conic, not only a circle)) and $P\to R$ (from line $AB$ to line $BC$). We must now show that $C\to C$. In this case $Q=C\Rightarrow L=AC\Rightarrow P=A\Rightarrow PR=AC\Rightarrow R=AC\cap BC=C$, and we're done.

grobber, i didn't understand your solution properly could u explain more?

Assume you have two lines $d_1,\ d_2$ and a point $X$ (which doesn't belong to $d_1$ or $d_2$). Take a variable line $d$ through $X$ which cuts $d_i$ in $A_i$. The function $f:d_1\to d_2,\ f(A_1)=A_2$ is called perspectivity and it'sa a particular case of function from $d_1$ to $d_2$ which preserves the cross-ratio of four points. If you have $g:d_1\to d_2$ s.t. $g$ preserves the cross-ratio, it can be shown (I won't show it here; try searching the Net for stuff like "homographic transformation", "perspectivity" etc.) that this is a perspectivity iff $g(O)=O$, where $O=d_1\cap d_2$. This is exactly what I did here.

oh! Grobber I know homographic transformation I had not understood the last part of your proof.

Oh.. Sorry for lecturing you Is it Ok now?

last line is'nt clear to me properly

What I'm trying to say is that according to the theorem I meantioned in one of my previous messages on this topic, in order to show that the transformation $Q\to R$ is a perspectivity from line $AC$ to line $BC$ (i.e. $QR$ passes through a fixed point), we must show that by this transformation the intersection point of the two lines $AC$ and $BC$ turns into itself. This means we want to prove that when $Q$ coincides with $C$, by constructing $R$, we also find it to coincide with $C$. That's why the last line begins with $Q=C$ and ends with $R=C$ (what we want to show is the implication $Q=C\Rightarrow R=C$).

Omid Hatami wrote: Suppose $T$ is intersection of $QR$ and $AB$ If the tangent from $T$ to circle (not $AB$ the other) is' not parallel to $AC$ by Brianchon theorem $PR$ , $AC$ and tangent from $T$ to circle are concurrent. But $PR$ is parallel to $AC$ contradiction. So $T$ is constant. I don't understand why we have $PR$ , $AC$ and tangent from $T$ to circle are concurrent by Brianchon theorem