Question

如圖，點(diǎn)P是⊙O 外一點(diǎn)，PA切⊙O于點(diǎn)A，AB是⊙O的直徑，連接OP，過(guò)點(diǎn)B作BC∥OP交⊙O于點(diǎn)C，連接AC交OP于點(diǎn)D．

(1)求證：PC是⊙O的切線；

(2)若PD=cm，AC=8cm，求圖中陰影部分的面積；

(3)在(2)的條件下，若點(diǎn)E是的中點(diǎn)，連接CE，求CE的長(zhǎng)．

   

Answer 1

證明： ⑴如圖，連接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的半徑，

  ∴PC是⊙O的切線． ·············································································································· 5分⑵解法一：

由（1）得PA，PC都為圓的切線，

∴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

由題意知OD為△ABC的中位線，

∴BC=2OD=6，AB=10． ······································································································· 8分

∴S_陰=S_半⊙O－S_△ACB=．

答：陰影部分的面積為．··················································································· 9分

解法二：

∵AB是⊙O的直徑，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_陰=S_半⊙O-S_△ACB=．

答：陰影部分的面積為．··················································································· 9分

（3）如圖，連接AE，BE_，過(guò)點(diǎn)B作BM⊥CE于點(diǎn)M．························································ 10分

∴∠CMB=∠EMB=∠AEB=90º，

又∵點(diǎn)E是的中點(diǎn)，

∴∠ECB=∠CBM=∠ABE=45º，CM=MB =，BE=ABcos45º=，······························· 11分

∴ EM=，

∴CE=CM+EM=．

答：CE的長(zhǎng)為cm． ····································································································· 12分