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Given triangle $ABC$ with the circumcircle $(O)$. An arbitrary line meets the sides $AC$, $AB$ at $D$, $E$, respectively, and meets $(O)$ at $P$ and $Q$. Let $BD$ meets $(O)$ at $M$, $CE$ meets $(O)$ at $N$. Let $I=MP\cap NQ$, $K=MN\cap PQ$. Prove that $AI\perp OK$.enter image description here

It's obviously that $I$ lies on the polar of $K$, hence, we just need to show that $A$ lies on the polar of $K$ (we may prove that $AK$ is the tangent line of $(O)$). I've tried to use Pascal theorem for some hexagon inscribed the circle (such as $AAMNPQ$) but got stuck at lots of points. I'm wondering whether we can apply harmonic division or any lemma relating pole and polar to solve this problem. I need help with this, thank you so much!

Steven Tran
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1 Answers1

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A quadrilateral any $MNPQ$ inscribed in a circle ($λ$) with center $O$ can have three points of intersection: $I = MP∩NQ$, $K = MN∩ PQ$ and $L = MQ∩NP$. We can prove (Cartesian coordinates) that point $I$ belongs to the radical line ($t$) referring to the circles (λ) and (ω) centered on the midpoint of $OK$. The Chordal Theorem guarantees that the radical line is perpendicular to the centers of the circles:$λ$ and $ω$. By the wording of the question, we have $MNPQ$. Taking vertex $A$ of triangle ABC as $A = λ∩t$, $D∈PQ$ (free point), $C = λ∩AD$ and $B = λ∩MD$ generate $MNPQ$.

quadrilaterl

Wilson Massaro
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