In the rhombus AVSD, the diagonals intersect at the point O. OM, OK, OE perpendiculars lowered to the sides AB

| November 24, 2020 | education

In the rhombus AVSD, the diagonals intersect at the point O. OM, OK, OE perpendiculars lowered to the sides AB, BC, SD, respectively. Prove that OM = OK, and find the sum of the angles MOB and COE.

Since a rhombus is a parallelogram in which all sides are equal, and its diagonals are mutually perpendicular and the intersection point is divided in half, then the AOB and BOS tracks are equal, which means that their heights drawn from equal angles will be equal.
Because AB || CD and OM are perpendicular to AB and OE are perpendicular to CD, then they lie on one straight line. Since the angle COE = Angle MOA and the angle MOB = angle DOE (as vertical) and the diagonals of the rhombus are mutually perpendicular, it turns out that the sum of the angles MOB and COE is 90 degrees.



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