As shown in the figure, in △ ABC, ﹥ ABC = ﹥ C, ﹥ EBC = ﹥ bed = 60 °, ad bisects ﹥ BAC, proving: ﹥ d = 30 °

As shown in the figure, in △ ABC, ﹥ ABC = ﹥ C, ﹥ EBC = ﹥ bed = 60 °, ad bisects ﹥ BAC, proving: ﹥ d = 30 °

It is proved that: as shown in the figure, extending ed and ad intersect BC with points g and f respectively, ∫ ABC = ∧ C, ∧ ABC is isosceles triangle, ∫ ad bisects ∧ BAC, ∨ AF ⊥ BC, that is, ∧ DFG = 90 °, ∧ EBC = ∧ bed = 60 °, ∧ DGF = 60 ° and ∧ EDA = ∧ GDF = 30 °

Take three sides of triangle ABC as equilateral triangles DBA, EBC and FAC on the same side of BC. (1) try to explain what quadrilateral afed is

A quadrilateral afed is a parallelogram
In triangle ABC and triangle FEC, from FC = AC, ∠ ACB = ∠ FCE (because ∠ ACB = 60 °～ ECA, ∠ FCE = 60 °～ ECA) and CB = CE, we can know the congruence of two triangles, and get EF = BC = DB = Da, ∠ CEF = ∠ CBA;
In the triangle DBE and triangle FEC, the congruence of the two triangles can be obtained by EC = be, ∠ EBD = ∠ CEF (∠ EBD = 60 °- ∠ EBA, ∠ CEF = ∠ CBA = 60 °- ∠ EBA), EF = dB, and de = FC = AF;
In quad afed, it is parallelogram by de = AF, EF = da

As shown in the figure, BD is divided into two parts, be is divided into ABC 2:5, DBE = 21 ° and the degree of ABC is calculated

Let ∠ Abe = 2x ° and get 2x + 21 = 5x-21. The solution is x = 14 and ﹥ ABC = 14 °× 7 = 98 °. The degree of ﹥ ABC is 98 degrees. So the answer is 98 degrees