一、选择题1．C 2．D 3．A 4．D 5．C
二、填空题6．110° 7．65°或115° 8．2 9．50°10． 24 cm
三、解答题11．解：∵AB是直径，BC与⊙O相切，∴AB上BC．∴∠ABC =90°∵∠C=25°，∴∠BOD=65°∴∠A =1/2∠BOD=32.5°14．（1）证明：连接OC∵ PD切⊙O于点C，∴ OC⊥PD∵BD⊥PD，∴OC∥BD∴∠OCB=∠CBD∵OC= OB，∴∠OCB=∠OBC∴∠CBD=∠OBC∴BC平分∠PBD（2）证明：连接AC∵AB是直径，∴∠ACB =90°=∠D∵∠CBD=∠OBC∴△ACB∽△CDB∴AB/BC=BC/BD∴BC²=AB ·BD．（3）解:∵∠PCA +∠ACO = 90°， ∠BCO +∠ACO =90°∴∠ PCA =∠OCB =∠CBO，∵∠P=∠P，∴△PA∽△PCB．∴PC/PB=PA/ PC∴PB=12∴ AB = PB - PA = 6．∴OC=3，PO=9∵OC∥BD∴∠POC = ∠PBD，∵∠P=∠P，∴△PCO∽△PDB，∴OC/BD=PO/PB∴BD =4
