Let ABCD be a parallelogram and p any point. Through P draw lines parallel to BC and AB to cut BA and CD in G and H, and AD and BC in E and F. Prove that the diagonal lines EG, HF, DB are concurrent.
I'm thinking about to build two coaxial triangle, so this triangle are copolar and the pole is where EG,HF and DB intersect.
Let GE extended cut FH extended at Q.
Extend BA to some point A’ (and C’, similarly defined on BC) such that the //gm BC’QA’ is formed.
$\angle DBC$ is the angle between the diagonal BD with the side BC of the //gm ABCD. $\angle QBC'$ is similarly defined wrt //gm A’BC’Q.
Since the two //gms are similar, we can say that $\angle DBC = \angle QBC'$. This further means BD is on BQ.