ABCD isosceles trapezoid, AC diagonal, AB = BC, ∠ CAD = 30 degrees. Find: ∠ BAC, ∠B, ∠BCD, ∠D.

Angle BCA = СAD as cross-lying angles at the intersection of parallel straight lines ABP and BC secant AC.
Since, by condition, AB = BC, then the triangle ABC is isosceles, then the angle BAC = BCA = 30.
Then the angle of ВСD = BAC + СAD = 30 + 30 = 60. The sum of the angles of the trapezoid at the lateral side is 180, then the angle ABC = 180 – 60 = 120.
In an isosceles trapezoid, the angles at the bases are equal, then the ВСD angle = ABC = 1200, the ADS angle = ВAD = 60.
Answer:: ∠ BAC = 30, ∠B = 120, ∠BCD = 120, ∠D = 60.