∵DP is the tangent of⊙ O.

∴OD⊥DP,∠ODP=90？

∴∠BDP=90？ -∠ODB

∵DC⊥OB,∴∠DCB=90？

∴∠BDC=90？ -∠OBD

∵OD=OB,∴∠ODB=∠OBD

∴∠BDP=∠BDC, that is, BD shares ∝∞∠CDP.

(2) If b is BH⊥DP in H, then BH=BC.

∵∠P=∠P,∴Rt△PBH∽Rt△PDC

∴BH/DC=PB/PD

∫tanP = DC/PC = 3/4，DC=6。

∴PC=8, PD of Pythagorean Theorem =10.

BH/6=(8-BC)/ 10

BH=BC=3