因为是正方体所以A1B1C1D1是正方形所以A1C1⊥B1D1,又棱边BB1⊥面A1B1C1D1,A1C1属于面A1B1C1D1,所以BB1⊥A1C1,又B1D1交BB1于B1,所以A1C1⊥面B1D1DB,又A1C1属于面A1C1CA,所以平面A1C1CA⊥平面B1D1DB
------------------一道证明题,不难,拜托了-------------------
因为是正方体所以A1B1C1D1是正方形所以A1C1⊥B1D1,又棱边BB1⊥面A1B1C1D1,A1C1属于面A1B1C1D1,所以BB1⊥A1C1,又B1D1交BB1于B1,所以A1C1⊥面B1D1DB,又A1C1属于面A1C1CA,所以平面A1C1CA⊥平面B1D1DB