Problem
GEO-B2-M08-P015 Independence of Circle Choice
#15
★★★★☆ Level 4 of 5
In a complete quadrilateral, point \(M_1\) is the second intersection of circles \((ABE)\) and \((ADF)\), while \(M_2\) is the second intersection of \((BCF)\) and \((CDE)\). Prove that \(M_1=M_2\).
Show that \(M_1\) lies on \((BCF)\) and \((CDE)\).
By Miquel's theorem, point \(M_1\), constructed as the second intersection of \((ABE)\) and \((ADF)\), also lies on \((BCF)\) and \((CDE)\). Therefore it is the common point of these two circles different from their already known common point \(C\). Hence it is \(M_2\).
Useful for confidently choosing any two circles.