Known: as shown in the figure, ∠ abd = ∠ DBC, ∠ ACD = ∠ DCE. (1) if ∠ a = 50 °, find the degree of ∠ D; (2) guess the relationship between ∠ D and ∠ a, and explain the reason; (3) if CD ‖ AB, judge the relationship between ∠ ABC and ∠ a

Known: as shown in the figure, ∠ abd = ∠ DBC, ∠ ACD = ∠ DCE. (1) if ∠ a = 50 °, find the degree of ∠ D; (2) guess the relationship between ∠ D and ∠ a, and explain the reason; (3) if CD ‖ AB, judge the relationship between ∠ ABC and ∠ a

(1) For △ BCD, ∠ DCE = ∠ DBC + ∠ D, and ∵ ∠ abd = ∠ DBC, ∵ ACD = ∠ DCE, ∵ ACD = ∠ abd + ∠ D. from the properties of the outer angle of the triangle, ∵ a + ∠ abd = ∠ D + ∠ ACD, can be solved by the above two formulas, ∵ d = 12 ∠ a; (2) from (1), ∵ d = 12 ∠ a; (3) if CD ‖ AB, ∵ ABC = ∠ DCE, and ∵ DBC + ∠ d = ∠ DCE, and ∵ d = 12 ∠ a, then 12 ∠ ABC + 1 2∠A=∠DCE，∴∠A=2∠DCE-∠ABC=∠ABC．