(1)试证:MN∥平面BCD;
(2)若∠BAC=∠CAD=∠DAB=90°,且AB=AD=1,AC=
,求证:BN⊥DM.
(1)试证:MN∥平面BCD;
(2)若∠BAC=∠CAD=∠DAB=90°,且AB=AD=1,AC=,求证:BN⊥DM.
证明:(1)设AM、AN分别交BC、CD的中点于E、F,
因为
=
=
+
=
+
,
所以
, 
,
共面,MN∥平面BCD.
(2)因为
=
-
=
(
+
)-
,
=
-
=
(
+
)-
,
所以
·
=
|AC|2-
(|AB|2+|AD|2)=
-
=0.
所以BN⊥DM.