Guess: ∠ADP=∠ABP

Proof idea: As shown in the figure, crossing P is the parallel line of AB and BC respectively.

First of all, we prove that △APH∽△CPF is equidiagonal, and get AH:CF=PH:PF.

Instead, it is the nature of parallelogram: the opposite sides are equal and the line segments are equal.

PE:BE=PH:HD, that is, PE:PH=BE:HD

It is easy to prove that ∠3=∠4

So △PHD∽△PEB

So ∠ADP=∠ABP