In rectangle ABCD O is the point of intersection of the diagonals, BH and DE are the heights of triangles ABO and COD

In rectangle ABCD O is the point of intersection of the diagonals, BH and DE are the heights of triangles ABO and COD, respectively, the angle BOH = 60 degrees, AH = 5 cm. Find BD

The AOB triangle is isosceles, the angles at the base of OAB and OBA are equal to 60 degrees, since the angle AOB = 60 and 120 degrees remain for the other two angles.
If the angle BOН = 60 degrees, then the angle HBO = 30 degrees, hence the angle HBA = 30 degrees.
We have: in the triangle AHB, by definition, ctgABH = BH / AH, which means BH = AH * ctgABH = 5 * ctg30 = 5 * √3 = 5√3 (cm).
In the BHO triangle, by definition, sinBOH = BH / OB, hence BO = BH / sinBHO = 5√3 / sin60 = 5√3 / √3 / 2 = 10 (cm).
So the diagonal of the rectangle BD = 2BO = 2 * 10 = 20 (cm).