MathDB
2009 JBMO Shortlist G1

Source: 2009 JBMO Shortlist G1

October 8, 2017
geometryJBMO

Problem Statement

Parallelogram ABCD{ABCD} is given with AC>BD{AC>BD}, and O{O} intersection point of AC{AC} and BD{BD}. Circle with center at O{O}and radius OA{OA} intersects extensions of AD{AD}and AB{AB}at points G{G} and L{L}, respectively. Let Z{Z} be intersection point of lines BD{BD}and GL{GL}. Prove that ZCA=90\angle ZCA={{90}^{{}^\circ }}.