解:(1)設(shè)∠BAC=x°,則∠DAC=2x°,
∴∠ABC=∠ACB=
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,
∠ABD=∠ADB=
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,
∠ACD=∠ADC=
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,
∴∠DBC=∠ABC-∠ABD,
=
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-
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,
=x°,
∠BDC=∠ADC-∠ADB,
=
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-
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,
=
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,
∴∠DBC/∠BDC=2;
(2)設(shè)∠BAC=x°,則∠DAC=3x°,
∴∠ABC=∠ACB=
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,
∠ABD=∠ADB=
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,
∠ACD=∠ADC=
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,
∴∠DBC=∠ABC-∠ABD,
=
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-
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,
=
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,
∠BDC=∠ADC-∠ADB,
=
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-
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,
=
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,
∴∠DBC/∠BDC=3;
(3)設(shè)∠BAC=x°,則∠DAC=nx°,
∴∠ABC=∠ACB=
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,
∠ABD=∠ADB=
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,
∠ACD=∠ADC=
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,
∴∠DBC=∠ABC-∠ABD,
=
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-
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,
=
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,
∠BDC=∠ADC-∠ADB,
=
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-
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,
=
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,
∴∠DBC/∠BDC=n.
故答案為:(1)2;(2)3;(3)n.
分析:(1)由題意,設(shè)∠BAC=x°,則∠DAC=2x°,∠DBC=∠ABC-∠ABD,∠BDC=∠ADC-∠ADB,根據(jù)三角形的內(nèi)角和定理,可得∠ABC=∠ACB=
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,∠ABD=∠ADB=
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,∠ACD=∠ADC=
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,代入即可求出;
(2)同理,當(dāng)∠DAC=3∠BAC時(shí),可求得∠DBC/∠BDC的值等于3;
(3)同理,當(dāng)∠DAC=n∠BAC時(shí),可求得∠DBC/∠BDC的值等于n.
點(diǎn)評:本題主要考查了等腰三角形的性質(zhì)和三角形的內(nèi)角和定理,由題意分別表示出各角的度數(shù),是解答本題的關(guān)鍵.