<Tri-2025-2191>, hint

Triangles HBC and HC'B' are similar.

Let <BHM=x=B'HN then <C'HM=A-x=<CHN

PH=HC' /(cosx cos(A-x)), QH=HB' /(cosx cos(A-x))

Hence Triangles HB'C' HPQ, HBC are similar PQ//BC.

Let D be the midpoint of BC. Locus of the midpoints of PQ is HD( except H, D )