Consider △CPB and △BPM,
- ∠CPM = ∠BPM = 90º (given)
- ∠PMB = 180º – ∠BPM – ∠PBM = 90º – ∠PBM
∠PBC = ∠MBC – ∠PMB = 90º – ∠PBM
∴, ∠PMB = ∠PBC
Therefore, △CPB ∼ △BPM (A.A.A. ∼).
Now, consider △CPD and △BPN,
- ∠DCP = ∠DCB – ∠PCB = 90º – ∠PBM
∠NBP = ∠NBM – ∠PBM = 90º – ∠PBM
∴, ∠DCP = ∠NBP - As MB = NB (given) and DC=BC (sides of square),
BN/CD = MB/BC
Also, PB/PC = MB/BC ( △CPB ∼ △BPM )
So, BN/CD = PB/PC
Therefore, △CPD ∼ △BPN (2 sides are proportional with equal inscribed angle).
As a result, ∠DPC = ∠NPB (properties of similar triangles).
∠CPB = 90º
⇒ ∠NPB + ∠CPN = 90º
⇒ ∠DPC + ∠CPN = 90º
⇒ ∠DPN = 90º
i.e. PD is perpendicular to PN.