In △ABC 1, d and k are the midpoint, ∴DK∥BC 1, (4 points).
DK again? Plane DCA 1, BC 1? DCA 1 plane, ∴bc 1∥DCA 1 plane, (6 points)
Figure 1, Figure 2 and Figure 3
(2) Proof: (Method 1) As shown in Figure 2, ac = bc, D is the midpoint of AB, ∴CD⊥AB.
And CD⊥DA 1, AB∩DA 1=D, CD ⊥ plane ABB 1A 1, (8 points).
Take the midpoint e and d of A 1B 1 as the midpoint of AB. ∴DE, BB 1 and CC 1 are parallel and equal.
∴DCC 1E is a parallelogram, ∴C 1E and CD are parallel and equal.
And CD⊥ plane ABB 1A 1, ∴C 1E⊥ plane ABB 1A 1, ∴ EBC 1, that is, the angle.
From the previous proof, we can see that the CD⊥ plane ABB 1A 1, CD ⊥ BB 1,
And AB⊥BB 1, AB∩CD=D, ∴BB 1⊥ plane ABC, ∴ this prism is a right-angled prism.
Let AC=BC=BB 1=2, ∴ BC 1 = 22, EC 1 = 2, ∠ EBC 1 = 30, (12 points).
(Method 2) As shown in Figure 3, ac = bc, D is the midpoint of AB, ∴CD⊥AB,
And CD⊥DA 1, AB∩DA 1=D, CD ⊥ plane ABB 1A 1, (8 points).
If we take the midpoint f of DA 1, KF∑CD, ∴KF⊥ plane abb 1a 1.
∴∠KDF is the angle formed by BC 1 and the plane ABB 1A 1 (10)
From the previous proof, we can see that the CD⊥ plane ABB 1A 1, CD ⊥ BB 1,
And AB⊥BB 1, AB∩CD=D, ∴BB 1⊥ plane ABC, ∴ this prism is a right-angled prism.
Let AC=BC=BB 1=2, ∴ KF = 22, DK = 2, ∴ KDF = 30, (12 points).