如图，点P是⊙O 外一点，PA切⊙O于点A，AB是⊙O的直径，连接OP，过点B作BC∥OP交⊙O于点C，连接AC交OP于点D．
(1)求证：PC是⊙O的切线；
(2)若PD=
cm，AC=8cm，求图中阴影部分的面积；
(3)在(2)的条件下，若点E是
的中点，连接CE，求CE的长．
如图，点P是⊙O 外一点，PA切⊙O于点A，AB是⊙O的直径，连接OP，过点B作BC∥OP交⊙O于点C，连接AC交OP于点D．
(1)求证：PC是⊙O的切线；
(2)若PD=cm，AC=8cm，求图中阴影部分的面积；
(3)在(2)的条件下，若点E是的中点，连接CE，求CE的长．
证明： ⑴如图，连接
OC，
∵PA切⊙O于A．
∴∠PAO=90º． ····················································································································· 1分
∵OP∥BC，
∴∠AOP=∠OBC，∠COP=∠OCB．
∵OC=OB，
∴∠OBC=∠OCB，
∴∠AOP=∠COP． ··············································································································· 3分
又∵OA=OC，OP=OP，
∴△PAO≌△PCO (SAS)．
∴∠PAO=∠PCO=90 º，
又∵OC是⊙O的半径，
∴PC是⊙O的切线． ·············································································································· 5分⑵解法一：
由（1）得PA，PC都为圆的切线，
∴PA=PC，OP平分∠APC，∠ADO=∠PAO=90 º，
∴∠PAD+∠DAO=∠DAO+∠AOD，
∴∠PAD =∠AOD，
∴△ADO∽△PDA． ············································································································· 6分
∴
，
∴
，
∵AC=8， PD=
，
∴AD=
AC=4，OD=3，AO=5，····························································································· 7分
由题意知OD为△ABC的中位线，
∴BC=2OD=6，AB=10． ······································································································· 8分
∴S阴=S半⊙O－S△ACB=
．
答：阴影部分的面积为
．··················································································· 9分
解法二：
∵AB是⊙O的直径，OP∥BC，
∴∠PDC=∠ACB=90º．
∵∠PCO=
90 º，
∴∠PCD+∠ACO=∠ACO+∠OCB=90 º，
即∠PCD=∠OCB．
又∵∠OBC =∠OCB，
∴∠PCD=∠OBC，
∴△PDC∽△ACB， ······································ 6分
∴
．
又∵AC=8， PD=，
∴AD=DC=4，PC=
．··················································································· 7分
∴
，
∴CB=6，AB=10， ················································································································ 8分
∴S阴=S半⊙O-S△ACB=．
答：阴影部分的面积为．··················································································· 9分
（3）如图，连接AE，BE，过点B作BM⊥CE于点M．························································ 10分
∴∠CMB=∠EMB=∠AEB=90º，
又∵点E是的中点，
∴∠ECB=∠CBM=∠ABE=45º，CM=MB =
，BE=ABcos45º=
，······························· 11分
∴ EM=
，
∴CE=CM+EM=
．
答：CE的长为cm． ····································································································· 12分