题目内容
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求证:(l)O1A2=O1D•O1C; (2)BE平分∠ABC.
分析:(1)由01A=O1B,根据等弧所对的圆周角相等,即可求得∠O1AB=∠O1CA,又由∠AO1C=∠DO1A,则可证得△AO1C∽△DO1A,根据相似三角形的对应边成比例,即可求得O1A2=O1D•O1C;
(2)由∠ADO1=∠CDB,∠O1AB=∠O1CB,易得∠AO1D=∠ABC,又由同弧所对的圆周角等于它对圆心角的一半,即可求得∠ABC=2∠ABE,则可得BE平分∠ABC.
(2)由∠ADO1=∠CDB,∠O1AB=∠O1CB,易得∠AO1D=∠ABC,又由同弧所对的圆周角等于它对圆心角的一半,即可求得∠ABC=2∠ABE,则可得BE平分∠ABC.
解答:证明:(1)∵01A=O1B,
∴∠ACO1=∠BCO1,
∵∠O1AB=∠O1CB,
∴∠O1AB=∠O1CA,
∵∠AO1C=∠DO1A,
∴△AO1C∽△DO1A,
∴
=
,
∴O1A2=O1D•O1C;
(2)∵∠ADO1=∠CDB,∠O1AB=∠O1CB,
又∵∠AO1D=180°-∠ADO1-∠O1AB,∠ABC=180°-∠CDB-∠O1CB,
∴∠AO1D=∠ABC,
∵∠AO1D=2∠ABE,
∴∠ABC=2∠ABE,
∴BE平分∠ABC.
∴∠ACO1=∠BCO1,
∵∠O1AB=∠O1CB,
∴∠O1AB=∠O1CA,
∵∠AO1C=∠DO1A,
∴△AO1C∽△DO1A,
∴
O1A |
O1D |
O1C |
O1A |
∴O1A2=O1D•O1C;
(2)∵∠ADO1=∠CDB,∠O1AB=∠O1CB,
又∵∠AO1D=180°-∠ADO1-∠O1AB,∠ABC=180°-∠CDB-∠O1CB,
∴∠AO1D=∠ABC,
∵∠AO1D=2∠ABE,
∴∠ABC=2∠ABE,
∴BE平分∠ABC.
点评:此题考查了相似三角形的判定与性质,圆周角与圆心角的性质等知识.此题综合性较强,图形较复杂,但难度适中,解题的关键是注意数形结合思想的应用.
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