Problem
GEO-B2-M01-P021 The Converse of Reim's Theorem
Two circles intersect at \(A\) and \(B\). A line through \(A\) meets the first circle again at \(C\) and the second again at \(D\). Point \(E\) lies on the first circle, and point \(F\) lies on the second. It is known that \(CE\parallel DF\). Prove that points \(E,B,F\) are collinear.
Use cyclicity to express angles \(\angle(CE,EB)\) and \(\angle(DF,FB)\), then use \(CE\parallel DF\).
Since \(A,B,C,E\) are cyclic, \(\angle(CE,EB)\equiv\angle(CA,AB)\). Since \(A,B,D,F\) are cyclic, \(\angle(DF,FB)\equiv\angle(DA,AB)\). But \(C,A,D\) are collinear, so the right-hand sides are equal. Therefore \(\angle(CE,EB)\equiv\angle(DF,FB)\). Since \(CE\parallel DF\), we get \(\angle(DF,EB)\equiv\angle(DF,FB)\), hence \(EB\parallel FB\). Both lines pass through \(B\), so \(E,B,F\) are collinear.
A strong problem: the student must reverse the proof of Reim's theorem.