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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 $(C)$

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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.

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  • $\begingroup$ Is the solution based on poles and polar lines interesting. I have such one. $\endgroup$ – Oldboy Jan 4 at 15:57
  • $\begingroup$ 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 $\endgroup$ – user600785 Jan 5 at 7:01
  • $\begingroup$ I don not have a solution based in angle chasing,but it is not based on pole and polar lines as well, you interested? $\endgroup$ – StAKmod Feb 18 at 0:53
  • $\begingroup$ @StAKmod lemme see and thanks $\endgroup$ – user600785 Feb 23 at 13:14

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