Lemma proposition_42B : forall B C D E J K R a b c e, Triangle a b c -> Midpoint b e c -> nCol J D K -> Midpoint B E C -> Cong E C e c -> nCol R E C -> exists X Z, PG X E C Z /\ EF a b e c X E C Z /\ CongA C E X J D K /\ OS R X E C.