Since $MP \parallel QH$ (both perpendicular to OA) and $MQ \parallel PH$ (both perpendicular to OB), the quadrilateral MPHQ is a parallelogram and its diagonals MH, PQ cut each other in half. This means that the point H is symmetrical to the point M with respect to the midpont of the segment PQ. According to the solution of the problem symmetric + locus, the locus of H is the segment DE, where D, E are the feet of the A- and B-altitudes of the triangle $\triangle OAB$. If the point M is inside of the triangle $\triangle OAB$, then a line through M parallel to AB meets the rays OA, OB at points A', B' and the locus of H for the point M on A'B' is the segment D'E' where D', E' are the feet of the A'- and B'-altitudes of the triangle $\triangle OA'B'$. The segments D'E' for all possible segments A'B' fill the triangle $\triangle ODE$.