We know that a triangle with two equal sides is called an isosceles triangle. Similarly, we define that at least one set of quadrilaterals with equal opposite sides is called an isosceles quadrilateral. (1) please write down the name of a special quadrilateral you have learned that is an isosceles quadrilateral; (2) as shown in the figure, in △ ABC, points D and E are on AB and AC respectively. Let CD and be intersect at point O, if ∠ A = 60 °, DCB = ∠ EBC = 12 ∠ A. please write an angle equal to ∠ a in the graph, and guess which quadrilateral in the graph is an equal opposite quadrilateral; (3) in △ ABC, if ∠ A is an acute angle not equal to 60 °, points D and E are on AB and AC respectively, and ∠ DCB = ∠ EBC = 12 ∠ A. explore whether there is an equal opposite quadrilateral in the graph satisfying the above conditions, and prove your conclusion

We know that a triangle with two equal sides is called an isosceles triangle. Similarly, we define that at least one set of quadrilaterals with equal opposite sides is called an isosceles quadrilateral. (1) please write down the name of a special quadrilateral you have learned that is an isosceles quadrilateral; (2) as shown in the figure, in △ ABC, points D and E are on AB and AC respectively. Let CD and be intersect at point O, if ∠ A = 60 °, DCB = ∠ EBC = 12 ∠ A. please write an angle equal to ∠ a in the graph, and guess which quadrilateral in the graph is an equal opposite quadrilateral; (3) in △ ABC, if ∠ A is an acute angle not equal to 60 °, points D and E are on AB and AC respectively, and ∠ DCB = ∠ EBC = 12 ∠ A. explore whether there is an equal opposite quadrilateral in the graph satisfying the above conditions, and prove your conclusion

(1) For example: parallelogram, isosceles trapezoid, etc. (2) answer: the angle equal to ∠ A is ∠ BOD (or ∠ COE), ∫ BOD = ∠ OBC + ∠ OCB = 30 ° + 30 ° = 60 °, ∫ a = ∠ BOD, conjecture: quadrilateral DBCE is equiopposite quadrilateral; (3) answer: at this time, there is equiopposite quadrilateral, which is tetragonal DBCE. Proof 1: as shown in the figure, make CG ⊥ be at point G, BF ⊥ CD intersect CD extension at point F As the public side, BC is the public side, BC is the public side, BC is the public side, BC is the public side, BC is the public side, BC is the public side, BC is the public side, BC is the public side, the public side, the public side, the public side, the public side, the public side, the △ BCF 8780; CBG, the \8780; CBG, \ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\deltabdc and In △ CFB, ∠ DBC = ∠ fcbbc = CB ∠ DCB = ∠ EBC ≠ △ BDC ≌ △ CFB (ASA), 〈 BD = CF, ∠ BDC = ∠ CFB, ≠ ADC = ∠ CFE, ∫ ADC = ∠ DCB + ∠ EBC + ∠ Abe, ∫ FEC = ∠ a + ∠ Abe, ∫ ADC = ∠ FEC, ≌} FEC = ∠ CFE, ∫ CF = CE, ∫ BD = CE, ∫ DBCE are equilateral quadrilateral. Note: when AB = AC, BD = CE still holds