<Qua-2025-779>, hint

Let AB=a, BC=b, <EBC=x, <FBA=y

since <AQF=<EBC=x, <EPC=<ABF=y, triangles BFQ and PEB are similar

EC=BC tan x=btanx, CP=CE/tany, CP=b tanx/ tany . CD/CP= a tany/ btanx.

Similarly AQ/AD=a tany/ b tanx

.Hence triangles DCP and QAD are similar and <QDA+<PDC=90