# 2004 IMO Shortlist Problems/G8

## Problem

A cyclic quadrilateral $ABCD$ is given. The lines $AD$ and $BC$ intersect at $E$, with $C$ between $B$ and $E$; the diagonals $AC$ and $BD$ intersect at $F$. Let $M$ be the midpoint of the side $CD$, and let $N \neq M$ be a point on the circumcircle of $\triangle ABM$ such that $\frac{AN}{BN} = \frac{AM}{BM}$. Prove that $E, F, N$ are collinear.

## Solution

Let $P = CD \cap EF$. Let $Q = AB \cap CD$. Let $R = AB \cap EF$. Let$(ABM)$ denote the circumcircle of $\triangle ABM$. Let $N' = EF \cap (ABM)$. Note that $N' = PR \cap (ABM)$

Claim: $P$ is on $(ABM)$. Proof: $(C, D; P, Q) = -1$ as complete quadrilaterals induce harmonic bundles. $QP \cdot QM = QC \cdot QD$ by Lemma 9.17 on Euclidean Geometry in Maths Olympiad. By power of a point theorem, $QP \cdot QM = QA \cdot QB$ and this is equivalent to our original claim.

$(A, B; R, Q) = -1$ as complete quadrilaterals induce harmonic bundles. By a projection through $P$ from $AQ$ onto $(ABM)$, $(A, B; N', M) = -1$. Since $\frac{AN}{BN} = \frac{AM}{BM}$, $M$ and $N$ are on the intersections of $(ABM)$ and an Appollonian circle centered on AB, so N and M are on the opposite sides of AB. Therefore, $(A, B; N, M) = -\frac{AN}{BN}\div\frac{AM}{BM} = -1$. By uniqueness of harmonic conjugate, $N = N'$