∴∠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.