Proof of Lemma 1 (2).     If  R is the length of the radius of K then the radius of both Feuerbach circles is R/2. On the other hand the Feuerbach circle F of ABC passes through M , midpoint of BC, through H, orthogonal projection of A on BC, and has its center above BC. Analogously for F', Feuerbach circle of A'B'C'. This makes obvious that F' is obtained from F by a parallel translation corresponding to v.