6. 如图,在正方体ABCD−A1B1C1D1ABCD-A_{1}B_{1}C_{1}D_{1}ABCD−A1B1C1D1,对角线A1CA_{1}CA1C与平面BDC1BDC_{1}BDC1交于点OOO.ACACAC、BDBDBD交于点MMM、EEE为ABABAB的中点,FFF为AA1AA_{1}AA1的中点,求证:(1)C1C_{1}C1、OOO、MMM三点共线(2)EEE、CCC、D1D_{1}D1、FFF四点共面(3)CECECE、D1FD_{1}FD1F、DADADA三线共点.