∴∠2=∠3.

AB = AD，

∴∠ 1=∠3.

∴∠ 1=∠2.

∵ quadrilateral ABCD is trapezoidal,

AB=DC，∠C=60，

∴∠ 1=∠2=30 .

That is ∠ Abd = 30.

(2) prove that ≈C = 60, ∠ 1 = ∠ 2 = 30,

∴∠BDC=90。

∴BC=2CD.

(3)

Solution: AE ⊥ BD, AE= 1

∴AB=2,BE=3.

∴CD=2,BD=23.

∴S trapezoidal ABCD =12× 23×1+12× 2× 23 = 33.