The diagonals of the parallelogram ABCD intersect at point O and are equal

The diagonals of the parallelogram ABCD intersect at point O and are equal to each other. The AOD angle is 140 degrees. Find the ABD angle.

Since, by condition, the diagonals of the parallelogram are equal, such a parallelogram is a rectangle.

Then the diagonals at the intersection point are divided in half and AO = ОD = OC = ОВ.

Then the triangle AOD is isosceles and the angle ADO = DAO = (180 – 140) / 2 = 40/2 = 20.

Angle СВО = АDО as cross-lying angles at the intersection of parallel straight lines ВС and АD of secant ВD. Angle CBD = ADO = 20.

Angle ABC = 90, then angle ABD = 90 – CBD = 90 – 20 = 70.

Answer: Angle ABD = 70.