Problem
GEO-B2-M04-P022 A Miquel Configuration as a Spiral Hint
Points \(B,A,D\) are collinear, and points \(C,A,E\) are collinear. Circles \((ABC)\) and \((ADE)\) meet again at \(M\). Prove that if \(MB=MD\), then \(MC=ME\).
Compare triangles \(MBC\) and \(MDE\): one angle is already equal, and also use angles subtended by the lines through \(A\).
From the cyclicity of \(A,B,C,M\), we have \(\angle MBC=\angle MAC\). From the cyclicity of \(A,D,E,M\), we have \(\angle MDE=\angle MAE\). Since \(A,C,E\) are collinear, \(\angle MAC=\angle MAE\), so \(\angle MBC=\angle MDE\). Also, from the previous configuration, \(\angle BMC=\angle DME\). Therefore \(\triangle MBC \sim \triangle MDE\). By the condition \(MB=MD\), the similarity ratio is \(1\), so \(MC=ME\).
This is a strong preparatory problem: the student essentially sees a spiral similarity with centre \(M\).