⊙O的弦AB,CD交于点P,PA=4,PB=3,PC=6,AE切⊙O于点A,AE与CD的延长线交于点E,EA=2根号5 求PE的长
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![⊙O的弦AB,CD交于点P,PA=4,PB=3,PC=6,AE切⊙O于点A,AE与CD的延长线交于点E,EA=2根号5 求PE的长](/uploads/image/z/3722035-67-5.jpg?t=%E2%8A%99O%E7%9A%84%E5%BC%A6AB%2CCD%E4%BA%A4%E4%BA%8E%E7%82%B9P%2CPA%EF%BC%9D4%2CPB%EF%BC%9D3%2CPC%EF%BC%9D6%2CAE%E5%88%87%E2%8A%99O%E4%BA%8E%E7%82%B9A%2CAE%E4%B8%8ECD%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%A4%E4%BA%8E%E7%82%B9E%2CEA%EF%BC%9D2%E6%A0%B9%E5%8F%B75+%E6%B1%82PE%E7%9A%84%E9%95%BF)
⊙O的弦AB,CD交于点P,PA=4,PB=3,PC=6,AE切⊙O于点A,AE与CD的延长线交于点E,EA=2根号5 求PE的长
⊙O的弦AB,CD交于点P,PA=4,PB=3,PC=6,AE切⊙O于点A,AE与CD的延长线交于点E,EA=2根号5 求PE的长
⊙O的弦AB,CD交于点P,PA=4,PB=3,PC=6,AE切⊙O于点A,AE与CD的延长线交于点E,EA=2根号5 求PE的长
由割线定理有,PA*PB=PC*PD,所以,PD=PA*PB/PC=4*3/6=2.
由切割线定理有:AE^2=ED*EC,所以,20=ED*(DE+PD+PC)=ED(ED+8),
解得:ED=2,(ED=-10,不合题意,舍去)
所以,PE=ED+PD=2+2=4.