<Tri-2025-2219>, hint

Since  triangles ADB, CFB, CEA are  similar(AD/BD=x/y=CE/AE)

BD=yc (AB=c)  BF=ya (BC=a) and <BDF=<ABC, Thus BD/BF=c/a=AB/BC.

Hence triangles BDF, BAC are similar . Similarly triangles FEC , BAC are similar

Hence DF=yb=AE, FE=xc=AD (AD/BD=x/y=CE/AE). ADFE is a parallelogram