I think that this is a classic one. From Pascal's theorem the intersection $M$ of $AC$ and $BD$ lies on the same line with $R,S$ so $M\in l$ Again by Pascal on degenerate $AACBBD$ we get that the intersection $L$ of $CB,AD$ lies on $l$. Finally by Pascal on degenerate $CCADDB$ we get that the intersection of the tangents at points $C,D$ lies on $l$ and we are done.

First, note that Pascal's theorem applied for $DAPBCQ$ yields $DA \cap BC \in \ell$. Furthermore, Pascal's theorem for $DBBCAA$ gives the collinearity of $DB \cap CA$, $K$ and $BC \cap AD$, thus $DB \cap CA$ lies on $\ell$ from the first Pascal. Finally, the same Pascal, applied this time for $DDACCB$ yields $CC \cap DD \in \ell$, as $\ell$ is the line determined by $DA \cap CB$ and $AC \cap BD$. //didn't see silouan's post; other mods, do delete if you find this useless. EDIT: Haha, ok, something to make up a little for the vacuous content. Here's a nice and easy generalization. Generalization. Let $\Gamma$ be a circle and $\ell$ a line lying outside $\Gamma$. Let $K \in \ell$ and let $AB$ and $CD$ be two chords of $\Gamma$ passing through $K$. Take $P$, $Q$ two points on $\Gamma$. Let $PA$, $PB$, $PC$, $PD$ meet $\ell$ at $X$, $Y$, $Z$, $T$, respectively, and then let $QX$, $QY$, $QZ$, $QT$ meet again $\Gamma$ at $R$, $S$, $U$, $V$. Then $RS$ and $UV$ meet on $\ell$. //didn't draw the diagram, so apologies if the conclusion should be reversed with the concurrency of the other pair of opposite sides; Pascal seems right though!

Funny problem, eh? Next time maybe, a problem where applying seven times Desargues theorem, on some special or degenerate cases (like parallelism, or other points at infinity) will yield some otherwise ridiculously complicated statement.

After we see that AD and BC meet on l, we easily get that l is the polar of X, where X is the intersection of AB and CD. By La Hire's theorem, the pole of CD must lie on l and we are done

Applying Pascal's theorem with $APBCQD$ we see that $BC \cap AD=Z \in l$.Let $AB \cap CD=J$.Then the polar of $K$ passes through $J \implies$ the polar of $J$ passes through $K$.Also by Brocard's theorem the polar of $J$ passes through $Z$.Thus the polar of $J$ is precisely $l$.In other words the tangents at $C$ and $D$ meet on $l$. Solution to Cosmin's generalization: Pascal's theorem on $UQRCPA \implies AU \cap RC \in l$ Pascal's theorem on $BPDSVQ \implies DS \cap BV \in l$ Pascal's theorem on $BRCDUA \implies BR \cap UD \in l$ Pascal's theorem on $SRBVUD \implies RS \cap UV \in l$ Nice generalization,Cosmin

sayantanchakraborty wrote: Solution to Cosmin's generalization: Pascal's theorem on $UQRCPA \implies AU \cap RC \in l$ Pascal's theorem on $BPDSVQ \implies DS \cap BV \in l$ Pascal's theorem on $BRCDUA \implies BR \cap UD \in l$ Pascal's theorem on $SRBVUD \implies RS \cap UV \in l$ Nice generalization,Cosmin My solution is the same so I share my diagram: There are many intersections on $\ell$ amazing problem!

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Drytime wrote: Let $\gamma$ be a circle and $l$ a line in its plane. Let $K$ be a point on $l$, located outside of $\gamma$. Let $KA$ and $KB$ be the tangents from $K$ to $\gamma$, where $A$ and $B$ are distinct points on $\gamma$. Let $P$ and $Q$ be two points on $\gamma$. Lines $PA$ and $PB$ intersect line $l$ in two points $R$ and respectively $S$. Lines $QR$ and $QS$ intersect the second time circle $\gamma$ in points $C$ and $D$. Prove that the tangents from $C$ and $D$ to $\gamma$ are concurrent on line $l$. Note that $\boxed{AA \cap BB = K}-(1)$, $\boxed{PA \cap QC = R}-(2)$ and $\boxed{PB \cap DQ = S}-(3)$ and let $\boxed{CC \cap DD = X}-(4)$. Now keeping in mind the pattern of intersections of the lines, in the $2^{\text{nd}}$ and the $3^\text{rd}$ box, we consider the hexagon $BPADQC$ and apply Pascal's theorem to it; on doing so we get the intersections: $PA \cap QC = R$, $\boxed{AD \cap CB = V}-(5)$(we may say) and $PB \cap DQ = S$ to be collinear. Hence $R, S, V$ are collinear and as, $\overline{RS} \equiv l \implies V$ lies on $l$. Now again after observing the pattern of intersections, of the lines in the $1^{\text{st}}$ and $5^{\text{th}}$ box, we consider the hexagon $AACBBD$ and again apply Pascal's theorem to get the intersections: $AA \cap BB = K, \boxed{AC \cap BD = U}-(6)$(we may say) and $CB \cap DA = V$ to be collinear. Hence $U, V, K$ are collinear and as $\overline{VK} \equiv l \implies U$ lies on $l$. Now for the last time we apply Pascal's theorem (as usual after observing the pattern of intersections, in the $5^{\text{th}}$ and the $6^{\text{th}}$ box) to the hexagon $ACCBDD$; on doing so, we get the intersections: $AC \cap BD = U, CC \cap DD = X$ and $CB \cap DA = V$ to be collinear. So, $U, V, X$ are collinear and again as $\overline{UV} \equiv l \implies X$ lies on $l$ and thus the tangents from $C$ and $D$ to $\gamma$ are concurrent on line $l$. $\blacksquare$

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It's Really Trivial... Let $K$= $ CC \cap DD$ , now as $AB \cap CD$ lies on $CD$ , which is polar of $K$ , so by La-Hire's $K$ lies on polar of $AB \cap CD$ which is $l$ by Brokard's Theorem.