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Problem: Given a point $P$ outside a circle $\Gamma$. A line passing through $P$ intersects the circle $\Gamma$ at points $A$ and $B$. Another line passing through $P$ intersects the circle $\Gamma$ at points $C, D$. $A \in \overline{PB}$, $C \in \overline{PD}$. The tangents of $\Gamma$ at $A$ and at $C$ intersect at point $Q$. The tangents of $\Gamma$ at $B$ and at $D$ intersect at point $S$. $AD$ and $BC$ intersect at point $R$. Prove that the points $P, Q, R, S$ are collinear.

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My Solution: In the degenerate cyclic hexagon $AACCDB$, if we apply Pascal's theorem we get that $P, Q, R$ are collinear. In the degenerate cyclic hexagon $ACDDBB$, if we apply Pascal's theorem we get that $P, R, S$ are collinear. Thus, four points $P, Q, R, S$ are collinear.

My question: I'm wondering how we can solve the problem in more fundamental ways, without using Pascal's theorem. Thanks for your ideas and advice.

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    $\begingroup$ This is truly a beautiful application of the Theorem of Pascal, i took a second look at it, and maybe it is useful to note that the picture may be "completed" with the same property to also cover the other intersections, say $X=AC\cap BD$, $Y=QA\cap SB$, and $Z=QC\cap SD$. Then we have the other line $XYRZ$ in the picture. At any rate, plus one! $\endgroup$
    – dan_fulea
    Commented Mar 15, 2023 at 16:39

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In the following proof, I will ignore the point S. This is because if PQR can be shown as a straight line, PSR will also be a straight line via the similar logic.

Let O be the center of circle ABDC (in red). It should be clear that AQCO (in green) is cyclic.

enter image description here

It should also be clear that all green marked angles are equal.

Let the circle that passes through ABR be $\omega$. Extend QR to cut $\omega$ at T. Since ABTR is cyclic, $\angle 7$ can therefore be marked as green. Further, this means T is also a con-cyclic point of AQCO. Then $\angle 8$ should also be green marked. “$\angle 8 = \angle 3$” implies CDTR (in blue) is cyclic.

$\angle BAT = \angle BRT = \angle TDC$” implies ATDP is also cyclic (in yellow). Since A, T, D, P are four con-cyclic points and $\angle ATQ = \angle 7 = \angle 3$, we can conclude that TRQP is a straight line.

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The given solution is wonderful, but ok, an alternative is needed. Here is a possibility using the reciprocal of the Theorem of Menelaus.

Let $2x$, $2y$, $2z$, $2w$ be the measures of the arcs $\overset\frown{AB}$, $\overset\frown{BD}$, $\overset\frown{DC}$, $\overset\frown{CA}$. We construct $T=AQ\cap BC$. See the inserted picture.

MSE question 4658880 inscribed quadrilateral concurrence property

Then we have access to "all angles" from the picture in terms of $x,y,z,w$. Then applying the sine theorem, we can express proportions of segments from the picture in terms of $\sin$ values. So we compute (ignoring the signs): $$ \begin{aligned} \dfrac{PA}{PB} & = \dfrac{PA}{PD} \cdot \dfrac{PD}{PB} = \dfrac{\sin x}{\sin z}\cdot\dfrac{\sin (x+w)}{\sin (x+y)}\ , \\ \dfrac{RB}{RT} & = \dfrac{RB}{RA} \cdot \dfrac{RA}{RT} = \dfrac{\sin z}{\sin x}\cdot\dfrac{\sin (y-x)}{\sin (x+w)}\ , \\ \dfrac{QT}{QA} &= \dfrac{QT}{QC} = \dfrac{\sin (x+y)}{\sin (y-x)}\ , \\[3mm] \dfrac{PA}{PB} \cdot \dfrac{RB}{RT} \cdot \dfrac{QT}{QA} &= 1\ , \end{aligned} $$ so by the reciprocal of the Theorem of Menelaus applied for $\Delta ABT$ the points $P,Q,R$ are colinear. A similar argument shows that $P,S,R$ are collinear.

$\square$

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  • $\begingroup$ Your solution with tutorial ideas is very clear and has elementary methods. Thank you for taking your time. $\endgroup$
    – scarface
    Commented Mar 15, 2023 at 8:16
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Is this more elementary?

Suppose $AC$ and $BD$ meeet at $M$. Then $PR$ is polar line for $M$. But $AC$ is polar for $Q$ and thus $Q$ lies on polar for $M$. Same is true for $S$ and thus $QS$ is also polar line for $M$ and thus a conclusion.

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