Problem
GEO-B2-M10-P021 The Miquel Point of a Triangle Configuration
#21
★★★★★ Level 5 of 5
In triangle \(ABC\), points \(D\in BC\), \(E\in CA\), \(F\in AB\). Prove that circles \((AEF)\), \((BFD)\), \((CDE)\) have one common point.
Take the second intersection of circles \((AEF)\) and \((BFD)\), then prove that it lies on \((CDE)\).
Let circles \((AEF)\) and \((BFD)\) meet again at \(M\). We need to prove that \(C,D,E,M\) are cyclic. From \(M\in(BFD)\), \(\angle DMF=\angle DBF=\angle CBA\). From \(M\in(AEF)\), \(\angle FME=\angle FAE=\angle BAC\). Therefore \(\angle DME=\angle DMF+\angle FME=\angle CBA+\angle BAC=180^\circ-\angle BCA\). But \(\angle DCE=\angle BCA\). Thus \(\angle DME+\angle DCE=180^\circ\), and \(C,D,E,M\) lie on one circle.
This is a strong but accessible version of Miquel without quoting the theorem.