In rhombus ABCD, the diagonals meet at point O. OM, OK, OE perpendiculars dropped on sides AB, BC, CD

In rhombus ABCD, the diagonals meet at point O. OM, OK, OE perpendiculars dropped on sides AB, BC, CD, respectively. Prove that OM = OK and find the sum of the angles MOB and COE.

In a rhombus, all sides are equal, then AB = BC = СD = AD.

The diagonals of the rhombus at the intersection point are halved and intersect at right angles.

Then the angle AOB = BOC = 900, angle ABO = СВO. Triangles ABO and CBO are rectangular, in which AB = BC, angle ABO = CBO, then triangle ABO is equal to triangle CBO in hypotenuse and acute angles. Since the triangles ABO and CBO are equal, then the heights drawn from the right angle are equal. ОМ = OK, as required.

Similarly, triangle AOB is equal to triangle СOD. Let the angle ВOM = X0, then the angle AOM = (90 – X) 0. Angle DOE = MOВ, then the angle СOE = 90 – DOE = 90 – MOВ. Then COE + MOВ = 90.

Answer: The sum of the angles COE and MOB is 90.