A straight line is drawn through the intersection point O of the diagonals of the parallelogram ABCD

A straight line is drawn through the intersection point O of the diagonals of the parallelogram ABCD, intersecting the sides AB and CD at points E and F. Prove that BE = DF

Consider triangles BOE and DOF.

BO = DO since the diagonals of the parallelogram are halved at the point of their intersection.

Angle BOE = DOF as vertical angles at the intersection of straight lines BD and EF.

Angle OBE = ODF as criss-crossing angles at the intersection of parallel straight lines AB and CD of the secant BD.

Then the triangles BOE and DOF are equal in side and two adjacent angles, and then BE = DF, which was required to prove.