Question

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．

   

Answer 1

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