Two non-parallel equal chords AB and CD are drawn in a circle centered at O. Point M is the midpoint of chords AB

Two non-parallel equal chords AB and CD are drawn in a circle centered at O. Point M is the midpoint of chords AB and CD. Point M is the midpoint of chord AB, and point H is the midpoint of chord CD. Prove that the angles HMO = MHO

Let the chords be disjoint.
If the chords are equal in length, then, at whatever points of the circle they are drawn, each point of one chord AB (closest to each point of the second chord) and the second chord CD will be equidistant from the center of the circle, as in a mirror image. If you build an isosceles trapezoid ABCD, then the lines between these points will be parallel to the bases of the trapezoid.
Thus, the central points of the first and second chords, M and H, respectively, also equidistant from the center of the circle O, form an isosceles triangle HMO, or МНО, with the point O.
The angles at the base in an isosceles triangle are equal, which means that the HMO angle is equal to the МНО angle.
The proof is complete.