Problem (2021G1): Let $ABCD$ be a parallelogram with $AC=BC.$ A point $P$ is chosen on the extension of ray $AB$ past $B.$ The circumcircle of $ACD$ meets the segment $PD$ again at $Q.$ The circumcircle of triangle $APQ$ meets the segment $PC$ at $R.$ Prove that lines $CD,AQ,BR$ are concurrent.
AlphaGeometry formalization: c a b = iso_triangle; d = parallelogram a b c d; p = on_line a b; o1 = circumcenter a c d; q = on_line p d, on_circle o1 a; o2 = circumcenter a p q; r = on_line p c, on_circle o2 a; x = on_line c d, on_line a q ? coll x b r
HAGeo formalization: C A B = isos_triangle; D = parallelogram A B C; AB = line A B; P = on_line AB; DP = line D P; (ACD) = circle A C D; Q = intersection DP (ACD); CP = line C P; (APQ) = circle A P Q; R = intersection CP (APQ); CD = line C D; AQ = line A Q; BR = line B R; X = intersection CD AQ; Prove: collinear X B R
Difficulty score: 2.2