Given a parallelogram ABCD and 2 segments FH and EG intersecting the diagonal AC at point P and terminating on 2 opposite parallelogram sides respectively, plus segments HG and EF.

Pprove that triangles EFP and HGP are similar.

Could I get a hint on how to solve this problem?

If it could be proven that HG and EF are parallel or by the same token, that angle AFE=GHC, the proof would follow. I have been unable to make any headway.