解:(1)設∠BAC=x°,則∠DAC=2x°,
∴∠ABC=∠ACB=

,
∠ABD=∠ADB=

,
∠ACD=∠ADC=

,
∴∠DBC=∠ABC-∠ABD,
=

-

,
=x°,
∠BDC=∠ADC-∠ADB,
=

-

,
=

,
∴∠DBC/∠BDC=2;
(2)設∠BAC=x°,則∠DAC=3x°,
∴∠ABC=∠ACB=

,
∠ABD=∠ADB=

,
∠ACD=∠ADC=

,
∴∠DBC=∠ABC-∠ABD,
=

-

,
=

,
∠BDC=∠ADC-∠ADB,
=

-

,
=

,
∴∠DBC/∠BDC=3;
(3)設∠BAC=x°,則∠DAC=nx°,
∴∠ABC=∠ACB=

,
∠ABD=∠ADB=

,
∠ACD=∠ADC=

,
∴∠DBC=∠ABC-∠ABD,
=

-

,
=

,
∠BDC=∠ADC-∠ADB,
=

-

,
=

,
∴∠DBC/∠BDC=n.
故答案為:(1)2;(2)3;(3)n.
分析:(1)由題意,設∠BAC=x°,則∠DAC=2x°,∠DBC=∠ABC-∠ABD,∠BDC=∠ADC-∠ADB,根據三角形的內角和定理,可得∠ABC=∠ACB=

,∠ABD=∠ADB=

,∠ACD=∠ADC=

,代入即可求出;
(2)同理,當∠DAC=3∠BAC時,可求得∠DBC/∠BDC的值等于3;
(3)同理,當∠DAC=n∠BAC時,可求得∠DBC/∠BDC的值等于n.
點評:本題主要考查了等腰三角形的性質和三角形的內角和定理,由題意分別表示出各角的度數(shù),是解答本題的關鍵.