解法：
四边形BCDA
已知∠C = 90°
∠BAD = 90°
∠ABC = 120°
∴ ∠ADC = 360° -- ∠C -- ∠BAD -- ∠ABC
= 360° -- 90° -- 90° -- 120°
= 60°
设 CB 与 DA 延长线交于 点E
则∠EDC = ∠ADC = 60°
Rt△EDC
由 tan∠EDC = EC/CD 得：
EC = CD × tan∠EDC
= CD × tan 60°
= 12 × √3
= 12√3
Rt△EBA
∠EBA = 180° -- ∠ABC
= 180° -- 120°
= 60°
由 cos∠EBA = AB/EB 得：
EB = AB / cos∠EBA
= AB / cos60°
= 2 / (1/2)
= 4
∴ BC = EC -- EB
= 12√3 -- 4
解法二：四边形BCDA
已知∠C = 90°
∠BAD = 90°
∠ABC = 120°
∴ ∠ADC = 360° -- ∠C -- ∠BAD -- ∠ABC
= 360° -- 90° -- 90° -- 120°
= 60°
设 AB 与 DC 延长线交于 点F
则Rt△ADF ∠F = 90° -- ∠ADC
= 90° -- 60°
= 30°
Rt△BCF 设 BC = x
则 BF = BC / sin∠F
= BC / sin30°
= x / (1/2)
= 2x
FC = BC / tan∠F
= BC / tan30°
= x / (√3/3)
= √3 x
Rt△ADF 由 cos∠F = AF / FD 得：
cos30° = (BF + AB) / (FC + CD)
∴ √3/2 = (2x + 2) / (√3 x + 12)
∴ √3 × (√3 x + 12) = 2 × (2x + 2)
∴ 3x + 12√3 = 4x + 4
∴ x = 12√3 -- 4
则 BC = 12√3 -- 4
解法三：过点B作AD平行线交CD于点M过M作MN ⊥ AD于点N
易证 四边形ABMN矩形
∴ MN = AB = 2
Rt△MND
MD = MN / cos∠ADC
= MN / cos 60°
= 2 / (√3/2)
= 4√3 / 3
∴ MC = CD -- MD
= 12 --（4√3 / 3）
= (36 -- 4√3)/3
Rt△BMC
BC = MC × tan∠BMC
= MC × tan 60°
= [ (36 -- 4√3)/3 ] × √3
= (36√3 -- 12) / 3
= 12√3 -- 4
解法四：过点C作 CP ⊥ AD 于点P
再过点B 作 BQ ⊥ CP 于点Q
则 CP = CD × sin∠D
= 6√3
CQ = CP -- QP
= CP -- AB
= 6√3 -- 2
BC = CQ / sin∠CBD
= CQ / sin30°
= (6√3 -- 2) / (1/2)
= (6√3 -- 2) × 2
=12√3 -- 4
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