In trapezoid ABCD with bases BC and AD, diagonals are drawn. It is known that the angle BDA = 32 degrees

In trapezoid ABCD with bases BC and AD, diagonals are drawn. It is known that the angle BDA = 32 degrees and the angle CAD = 42 degrees. Determine the DBC angle and ACB angle.

Since ABCD is a trapezoid, BC is parallel to AD, straight lines AC and BD are secant, intersecting parallel lines.

The angles ACD and CAD are equal, as are the criss-crossing angles at the intersection of parallel lines of the secant AC, therefore, ACB = 42.

The angles are equal, as the angles lying crosswise at the intersection of parallel lines of the secant ВD, therefore, DBC = 32.

Answer: ACB = 42, DBC = 32.