Collinearity of $4$ points

Question

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.

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.

Answer 1

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.

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.

$\square$

Answer 2

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.

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.

Answer 3

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.