The diagonals of rectangle ABCD meet at point O. find the angles of the triangle COD if the angle ABD = 40 degrees.

All corners of the rectangle are 90 degrees. Angle abd and angle odc are criss-crossing angles with parallel lines ba and cd. By the property of the intersecting angles, they are equal, hence the angle cdp = 40 degrees. By the property of the diagonals of the rectangle at the intersection point, they are divided in half, and the diagonals are also equal. Consequently, consider the triangle cod, the sides oc and od are equal (from the property of the diagonals of a rectangle), and by the theorem on the angles of a triangle, the sum of all angles of a triangle is 180 degrees. Therefore, the angle cod = 180-40-40 = 100 degrees.
Answer: 40,40,100