(i) AC⊥BD, so BD is perpendicular to the A1AC plane. If BD is perpendicular to the A1AC plane, then BD is perpendicular to any segment of the plane, and BD⊥ A1C.
(ii) BD is a segment on the line of intersection of planes A1BD and BC1D. We also know that both A1E and C1E are perpendicular to BD, because they are part of the A1AC plane. This means that the angle between planes A1BD and BC1D is equal to ∠A1EC1.
∠A1EC1 = 180o - ∠A1EA - ∠C1EC.
We already know AA1 = CC1 = √3. In order to determine ∠A1EA and` ∠C1EC, we need to find AE and CE.
AD⊥DC so triangle ADC is right.
AC2 = AD2 + DC2 = 22 + (2√3)2 = 16
AC = 4
cos(∠CAD)= AD/AC = 1/2 so ∠CAD = 60o
cos(∠CAD)= AE/AD, so 1/2 = AE/AD. and AE = 1.
CE = AC - AE = 4 - 1 = 3.
tan(∠AEA1) = AA1/AE = (√3)/1 = √3, so ∠AEA1 = 60o.
tan(∠CEC1) = CC1/CE = (√3)/3, so ∠CEC1 = 30o.
∠A1EC1 = 180o - 60o - 30o = 90o.
The angle between planes A1BD and BC1D is 90o.
(iii) We draw a parallel to AD from point B.
AD and BP are parallel and ∠ADC is right, so ∠BPC is right.
Triangles ADC and ABC are similar, so ∠ACD = ∠ACB = 30o.
∠DCB = 2∠ACD = 60o.
In the right triangle BPC, BP = BC sin60o = 2√3((√3)/2) = 3.
Triangle BC1P is right because BP ⊥ plane CC1D1D.
BC12 = CC12 + BC2 = (√3)2 + (2√3)2 = 3 + 12 = 15.
BC1 = √15.
cos(∠PBC1)= BP/BC1 = 3/√15 = √(3/5).
∠PBC1 = arccos(√(3/5)).
In conclusion, the angle formed by lines AD and BC1 is equal to arccos(√(3/5)).