Problem
GEO-B2-M01-P015 Reim's Theorem
#15
★★★★☆ Level 4 of 5
Two circles intersect at \(A\) and \(B\). A line through \(A\) meets the first circle again at \(C\) and the second again at \(D\). A line through \(B\) meets the first circle again at \(E\) and the second again at \(F\). Prove that \(CE\parallel DF\).
Compare angles \(\angle(CE,EB)\) and \(\angle(DF,FB)\).
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 \(\angle(CA,AB)\equiv\angle(DA,AB)\). Also \(E,B,F\) are collinear. Therefore \(CE\) and \(DF\) form equal angles with the same line, so \(CE\parallel DF\).
This is already a genuine olympiad technique, but it is entirely based on angles.