Answer: \(\triangle APD \sim \triangle CPB\) (AA test), which gives \(\frac{AP}{PD} = \frac{PC}{BP}\).
In \(\triangle APD\) and \(\triangle CPB\):
- \(\angle PAD \cong \angle PCB\) (alternate angles, AD ∥ BC, AC is a transversal)
- \(\angle APD \cong \angle CPB\) (vertically opposite angles)
\(\therefore \triangle APD \sim \triangle CPB\) (AA test).
\[\frac{AP}{CP} = \frac{PD}{PB}\]Rearranging (alternendo),
\[\frac{AP}{PD} = \frac{PC}{BP}\]Hence shown.