<Tri-2025-2288>, hint

Since P, Q are the center of right triangles BYQ, CXP, BP=PY=PQ=QX= QC

Let M be the midpoint of BC, MG/GA=1/2=MQ/QC, thus GQ//AC, GQㅗPX and PQ=QC

. Triangle GPX is isosceles. Hence GP=GX, similarly GQ=GY. Hence triangles GQX and GYP are congruent.