ISL 2024 G4
Let ABCD be a quadrilateral with AB parallel to CD and AB < CD. Lines AD and BC intersect at a point P. Point
X ≠ C on the circumcircle of triangle ABC is such that PC = PX. Point Y ≠ D on the circumcircle of triangle ABD is such that PD = PY. Lines AX and BY intersect at Q.
Prove that PQ is parallel to AB.
Let l be the line parallel to AB passing through P
define: R = AC \cap BD,S = AC \cap l, T = BD \cap l note that (S,R;A,C) =-1
Let PX,PY intersect (ABC), (ABD) again at U,V respectively it is clear that the homothety that sends AB \rightarrow DC sends V \rightarrow Y, U \rightarrow X.
now let Q_1 be a point on l such that \triangle CUA \stackrel{+}{\sim} CPQ_1 we will define Q_2 similarly but for D. note that: \frac{Q_1A}{Q_1C} = \frac{PU}{PC} = \frac{PB}{PC} = \frac{RA}{RC} also \angle CPQ_1 = \angle CVA = \angle ABP so Q_1 \in l and Q_1R, Q_1S are internal/external angle bisectors of AQ_1C so from the harmonic bundle we get that Q_1 = Q_2 = foot of R on l
Finally:
\angle CAQ_1 = \angle CUP = 180 - \angle CAX \Rightarrow X,A,Q_1 are collinear and similarly for Y,B,Q_2 so Q_1 = Q_2 = Q and hence PQ \parallel AB
