Through the point O of the intersection of the diagonals of the parallelogram ABCD, a straight line is drawn that intersects

Through the point O of the intersection of the diagonals of the parallelogram ABCD, a straight line is drawn that intersects the sides AB and CD at points P and T, respectively. Prove that BP = DT.

Let us prove that the triangle BOP and DOT are equal.

Angle BOP = DОТ as vertical angles at the intersection of straight lines PT and BD. The angle ОDТ is equal to the angle ОВР as criss-crossing angles at the intersection of parallel lines AB and BC of the secant BD.

The segment OB = OD, since the diagonals of the parallelogram at the intersection point are divided in half. Then the triangle BOP = DOT along the side and two adjacent corners. Then BP = DТ, as required.