如图,已知⊙O1与⊙O2都过点A,⊙O1交O1O2于点B,连接AB并延长交⊙O2于点C,连接O2C,且O2C⊥O1O2(1)证:AO1是⊙O2的切线(2)证:AB×BC=2BO2×BO1.十万火急,
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 05:38:54
![如图,已知⊙O1与⊙O2都过点A,⊙O1交O1O2于点B,连接AB并延长交⊙O2于点C,连接O2C,且O2C⊥O1O2(1)证:AO1是⊙O2的切线(2)证:AB×BC=2BO2×BO1.十万火急,](/uploads/image/z/4064826-66-6.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5%E2%8A%99O1%E4%B8%8E%E2%8A%99O2%E9%83%BD%E8%BF%87%E7%82%B9A%2C%E2%8A%99O1%E4%BA%A4O1O2%E4%BA%8E%E7%82%B9B%2C%E8%BF%9E%E6%8E%A5AB%E5%B9%B6%E5%BB%B6%E9%95%BF%E4%BA%A4%E2%8A%99O2%E4%BA%8E%E7%82%B9C%2C%E8%BF%9E%E6%8E%A5O2C%2C%E4%B8%94O2C%E2%8A%A5O1O2%EF%BC%881%EF%BC%89%E8%AF%81%EF%BC%9AAO1%E6%98%AF%E2%8A%99O2%E7%9A%84%E5%88%87%E7%BA%BF%EF%BC%882%EF%BC%89%E8%AF%81%EF%BC%9AAB%C3%97BC%EF%BC%9D2BO2%C3%97BO1.%E5%8D%81%E4%B8%87%E7%81%AB%E6%80%A5%2C)
如图,已知⊙O1与⊙O2都过点A,⊙O1交O1O2于点B,连接AB并延长交⊙O2于点C,连接O2C,且O2C⊥O1O2(1)证:AO1是⊙O2的切线(2)证:AB×BC=2BO2×BO1.十万火急,
如图,已知⊙O1与⊙O2都过点A,⊙O1交O1O2于点B,连接AB并延长交⊙O2于点C,连接O2C,且O2C⊥O1O2(1)证:AO1是⊙O2的切线(2)证:AB×BC=2BO2×BO1.十万火急,
如图,已知⊙O1与⊙O2都过点A,⊙O1交O1O2于点B,连接AB并延长交⊙O2于点C,连接O2C,且O2C⊥O1O2(1)证:AO1是⊙O2的切线(2)证:AB×BC=2BO2×BO1.十万火急,
证明:
(1)
∠O2AB+∠O1AB
=∠O2CB+∠O1BA
=∠O2CB+∠O2BC
=180°-∠BO2C
=90°
∴O2A⊥O1A
∴AO1是⊙O2的切线
(2)
过O2做O2D⊥AC于D
AB×BC
=(AD-BD)(AD+BD)
=AD²-BD²
=AO2²-BO2²
=O1O2²-O1A²-BO2²
=O1O2²-O1B²-BO2²
=(O1B+O2B)²-O1B²-BO2²
=2BO2×BO2
证毕.
如仍有疑惑,欢迎追问.祝:
无图无解,带图来
△∵∴⊥∥∠≌∽√°π (1) ∵O2C⊥O1O2 ∴∠C+∠O2BC=90 ∵O1A=O1B=O1半径 ∴∠O1BA=∠O1AB 而∠O1BA和∠O2BC是对顶角,相等; ∴∠C+∠O1BC=90 而∠C=∠O2AC(半径所对角) ∴∠O2AB+∠O1BC=90 ∴AO1是⊙O2的切线 (2) 在△ADB和△O2BC中: 各有一直角,一对对顶角; ∴△ADB∽△O2BC 则对应边成比例,即AB/BO2=BD/BC 而BD是O1直径=2O1B, 代入及变形即可得AB×BC=2BO2×BO1