# Prove that $Q$, $S$ and $T$ lie on the same line.

Let $$(C)$$ be a circle , and $$ABCD$$ be a quadrilateral inscribed in $$(C)$$

Let $$P$$ be the intersection of $$(AD)$$ and $$(BC)$$. And $$Q$$ the intersection of $$(AB)$$ and $$(CD)$$

Let $$S$$ and $$T$$ be points in $$(C)$$ such that $$(PS)$$ and $$(PT)$$ are tangents to $$(C)$$

this problem can be done by Lahire theorem or projective geometry or polarisation.

But I wanna if there is a simple solution by angle chasing or radical axis

here is what I think we should do :

$$\widehat{OSP}=180-\widehat{OTP}=90$$ so OSPT is cyclic

so the problem reduces itself to prove that Q lies on the radical axis of $$(C)$$ and the circle where $$OSPT$$ is inscribed.

N.B: This problem is taken from a preparation test for IMO 2020 in Morocco.

• Is the solution based on poles and polar lines interesting. I have such one. – Oldboy Jan 4 at 15:57
• I aleady have a solution based on poles nd polar lines , but I'am looking for a an easy solution based in angle chasing or radical axis – user600785 Jan 5 at 7:01
• I don not have a solution based in angle chasing,but it is not based on pole and polar lines as well, you interested? – StAKmod Feb 18 at 0:53
• @StAKmod lemme see and thanks – user600785 Feb 23 at 13:14