Pass a as AE⊥BC and pass BC to E.

So EC=AD=3? AE=DC=4？ BE=√(25- 16)=3

∴BC=6

PM = MN OB = OP

∴∠COD=∠DOP？ ∠POH=∠HOB

∴∠COD+∠BOH=90

∠∠COD, ∠B is the complementary angle of∠∠ Boh.

∴∠COD=∠B？ AB∨DO？ Box=AD3

So OC=3