<Qua-2025-792>, hint

Let O', O" be the circumcenters of triangles ABC, ADC.

MN meet the midpoind of AC at Q.  let <ABC=X, <ADC=Y

O',O", Q are collinear and O'O" ㅗAC. Hence O'Q=(AC/2) cotX, O"Q=(AC/2) cotY

O'Q=O"Q---> X=Y, ABCD is parallelogram.