In the rhombus ABCD O is the point of intersection of the diagonals, E and F are the midpoints

In the rhombus ABCD O is the point of intersection of the diagonals, E and F are the midpoints of sides BC and DC. Prove that EF = BO and EF is perpendicular to AC.

Since, by condition, the points E and F are the midpoints of the segments BC and DS, the segment EF is the midline of the triangle BCD. The midline of a triangle is equal to half the length of the base of the triangle, to which it is parallel, then EF = BD / 2.

Since the diagonals of the rhombus at the point of their intersection are divided in half, then ОВ = ОD = ВD / 2, then EF = ОВ = ОD.

Since the middle line of the triangle is parallel to the base, then EF is parallel to BD, and since the diagonal of the rhombus is perpendicular, then EF is perpendicular to BD, which was required to prove.