<Tri-2026-2328>, hint
Suppose line AJ cut BC in D'. By Ceva's theorem with triangle ABC and BE,CF, AD'
concurring at J BD'/D'C=(s-b)/(s-c), 2s=a+b+c, Hence D'=D.
Menelaus theorem to triangle ABD with transversal F-J-C, JA/DJ=GK/KD={a(s-a)}/{(s-b)(s-c)}
Because AG//JD, both are perpendicular to EF
