如图,点O为△ABC的外心,点I为△ABC的内心,连接AI并延长交圆O于点D,连接BD,CD,BI
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如图,点O为△ABC的外心,点I为△ABC的内心,连接AI并延长交圆O于点D,连接BD,CD,BI
(1)求证 BD=DI=DC (2)连接IC,若角BAC=44°,试求角BIC的度数
![](http://img.wesiedu.com/upload/e/2e/e2ec0c358d0576eb3d101d1271eba30e.jpg)
(1)求证 BD=DI=DC (2)连接IC,若角BAC=44°,试求角BIC的度数
![](http://img.wesiedu.com/upload/e/2e/e2ec0c358d0576eb3d101d1271eba30e.jpg)
![如图,点O为△ABC的外心,点I为△ABC的内心,连接AI并延长交圆O于点D,连接BD,CD,BI](/uploads/image/z/647494-70-4.jpg?t=%E5%A6%82%E5%9B%BE%2C%E7%82%B9O%E4%B8%BA%E2%96%B3ABC%E7%9A%84%E5%A4%96%E5%BF%83%2C%E7%82%B9I%E4%B8%BA%E2%96%B3ABC%E7%9A%84%E5%86%85%E5%BF%83%2C%E8%BF%9E%E6%8E%A5AI%E5%B9%B6%E5%BB%B6%E9%95%BF%E4%BA%A4%E5%9C%86O%E4%BA%8E%E7%82%B9D%2C%E8%BF%9E%E6%8E%A5BD%2CCD%2CBI)
∵O为内心 ,∴∠DAB=∠DAC,∴弧BD=弧CD,∴BD=CD,
∵∠DBI=∠DBC+∠IBC=∠DAC+1/2∠ABC=1/2(∠BAC+∠ABC),
∠DIB=∠IBA+∠IAB=1/2(∠ABC+∠BAC),
∴∠DBI=∠DIB,∴BD=DI,
∴BD=DI=CD.
⑴∠BIC=∠DIB+∠DIC
=1/2∠ABC+1/2∠ABC+1/2∠ACB+1/2∠BAC
=1/2(∠ABC+∠BAC+∠ABC)+1/2∠BAC
=90°+1/2∠BAC
=112°.
∵∠DBI=∠DBC+∠IBC=∠DAC+1/2∠ABC=1/2(∠BAC+∠ABC),
∠DIB=∠IBA+∠IAB=1/2(∠ABC+∠BAC),
∴∠DBI=∠DIB,∴BD=DI,
∴BD=DI=CD.
⑴∠BIC=∠DIB+∠DIC
=1/2∠ABC+1/2∠ABC+1/2∠ACB+1/2∠BAC
=1/2(∠ABC+∠BAC+∠ABC)+1/2∠BAC
=90°+1/2∠BAC
=112°.
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