5. Connect HC, HP, AC and OH to AC in E; ∴∠ BHC = ∠ BPH,b =∠∴△BCH∠△BHP∴BC:BH = BH:BP ∴./BC=6? /4 = 9, CP = BP-BC = 9-4 = 5÷BH cuts ⊙O in H∴OH⊥BH, that is ∠ OHB = 90 ? AP is the diameter of ∴ O ∴ ACP. The radius PC = 2.5 ∴⊙ O OD=OE+ED=2.5+4=6.5.
5] Connect AB, AC, ∠ Bao = ∠ BCO = 60, ∠ AOC = ∠ AOB-∠ COB = 90-45 = 45 = ∠ ACO ∴ AC = AO = AB = ∠ 2. +AO? = √2ao = 4, ∴⊙O radius is √ 2, OC length is 4.