Problem
GEO-B2-M08-P008 The Fourth Circle
#8
★★★☆☆ Level 3 of 5
Under the conditions of the previous problem, prove that \(C,D,E,M\) lie on one circle.
Prove \(\angle CME=\angle CDE\) analogously to the previous proof.
From the previous problem we already know that \(M\in (BCF)\). Hence \(\angle CMB=\angle CFB\), which is the angle between lines \(l_4\) and \(l_2\). Also, from \(M\in (ABE)\) we get \(\angle BME=\angle BAE\), the angle between \(l_2\) and \(l_1\). Therefore \(\angle CME=\angle CMB+\angle BME\) is the angle between \(l_4\) and \(l_1\), that is \(\angle CDE\). By the converse cyclicity criterion, \(C,D,E,M\) lie on one circle.
This may be treated as the symmetric step after the previous problem.