Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 National olympiad Prove it

[Pascal's Theorem]

Quadrilateral $ABCD$ is inscribed in circle $S$ ; $X$ is an arbitrary point, $M$ and $N$ are the second points of intersection of lines $XA$ and $XD$ with circle $S$ . Lines $DC$ and $AX$ , $AB$ and $DX$ intersect at points $E$ and $F$ . Prove that the point of intersection of lines $MN$ and $EF$ lies on line $BC$ .

Solution

Let $K$ be the point of intersection of the lines $B C$ and $M N$ . Applying Pascal's theorem to the points $A, M, N, D, C, B$ , we get that the points $E, K$ and $F$ lie on the same line, which means that $K$ is the point of intersection of the lines $M N$ and $E F$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.