初中几何难题(圆)数学高手帮忙做一下。谢谢! 初中几何数学题,高手进,快!
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http://zhidao.baidu.com/question/276043239.html\uff1b
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\u53d6MN\u4e2d\u70b9P\uff0c\u8fdePO\u3002\u90a3\u4e48PO\u5e73\u884cAM\uff0c\u4e14PO=PM=PN
\u89d2AMN=60\u5ea6,\u89d2OPM=120\u5ea6\uff0c
\u89d2PMO=30\u5ea6\uff0c\u89d2AMO=30\u5ea6
\u4f5cOD\u5782\u76f4\u4e8eMN\u4e8eD\uff0c\u5219\u4e09\u89d2\u5f62OMA\u4e0eOMD\u5168\u7b49\u3002OD=OA
\u6240\u4ee5MN\u4e0e\u5706O\u76f8\u5207.
点I为⊿ABC的内心,则:∠CAI=(1/2)∠CAB=25°,∠ACI=(1/2)∠ACB=30°.
∴∠AIC=180°-(∠CAI+∠ACI)=180°-(25°+30°)=125°;
∵∠ADE=∠ABC=70°;DE=AD.
∴∠DEA=∠DAE=(180°-∠ADE)/2=(180°-70°)/2=55°.
则∠DEA+∠AIC=180°.故点A,I,C,E四点在同一个圆上.
∴∠DEI=∠CAI=25°.
已知E是圆内接四边形ABCD的边CD的延长线上一点,I是△ABC的内心,若∠ABC=70°,∠ACB=60°,DE=DA,则∠DEI的度数是( 25° )
连接AI CI
△ABC中
∠BAC=180°-∠ABC-∠ACB=180°-70°-60°=50°
∠1=∠ACB/2=30°
∠2=∠BAC/2=25°
△DAE中
ED=AD
∠DEA=∠DAE
∠DEA=∠DAE=(180°-70°)/2=55°
△DAE中
∠AIC=180°-∠1-∠2=125°
∠AIC+∠DEA=125°+55°=180°
AECI共圆
∠DEI=∠2=25°
35/2度
点I为⊿ABC的内心,则:∠CAI=(1/2)∠CAB=25°,∠ACI=(1/2)∠ACB=30°.
∴∠AIC=180°-(∠CAI+∠ACI)=180°-(25°+30°)=125°;
∵∠ADE=∠ABC=70°;DE=AD.
∴∠DEA=∠DAE=(180°-∠ADE)/2=(180°-70°)/2=55°.
则∠DEA+∠AIC=180°.故点A,I,C,E四点在同一个圆上.
∴∠DEI=∠CAI=25°.
∠DEI的度数是(35°)
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