Problem
GEO-B2-M04-P012 Two Circles Through One Point
#12
★★★☆☆ Level 3 of 5
Points \(B,A,D\) are collinear, and points \(C,A,E\) are collinear. Circles \((ABC)\) and \((ADE)\) meet again at \(M\). Prove that \(\angle BMC=\angle DME\).
Express both angles through the angle between the two lines passing through \(A\).
Since points \(A,B,C,M\) lie on one circle, \(\angle BMC=\angle BAC\). Since points \(A,D,E,M\) lie on one circle, \(\angle DME=\angle DAE\). But \(AB\) and \(AD\) are one line, and \(AC\) and \(AE\) are another line. Hence \(\angle BAC=\angle DAE\). Therefore \(\angle BMC=\angle DME\).
This is a gentle entrance into Miquel configurations and spiral similarity.