When the AP passes through the AP at point B', the vertical line segment BB' passes through B'; The vertical line segment EE passing through E intersects with AP as AP' at point E'

Then E'E//DO//B'B

From the theorem that parallel lines bisect line segments, DE'/DB' =1/3 is obtained; EE'/BB'= 1/3

∠DE'E=∠DB'B=90

So ⊿DE'E∽⊿DB'B

Then ∠ Aduh =∠PDB got the certificate.