It is known that: as shown in the figure, in isosceles trapezoid ABCD, ad ‖ BC, ab = DC, passing through point D as parallel line De of AC, intersecting the extension line of BA at point E

It is known that: as shown in the figure, in isosceles trapezoid ABCD, ad ‖ BC, ab = DC, passing through point D as parallel line De of AC, intersecting the extension line of BA at point E

It is proved that: (1) the ∵ quadrilateral ABCD is isosceles trapezoid, ∵ AC = dB (the two diagonals of isosceles trapezoid are equal), ∵ AB = DC (known), BC = CB (common edge), ∵ ABC ≌ DCB (SSS); (2) from (1), ≌ ABC ≌ DCB, ∤ ACB = ∠ DBC, ∤ ABC = ∠ DCB (the corresponding angles of congruent triangles are equal); ∥ ad ∥ BC (known), ≌ DAC = ∠ ACB (two straight lines) In this paper, we present a new method to solve the problem, that is, parallel, equal internal stagger angle, ead = ∠ ABC, ed AC, EDA = DAC, EDA = DBC, ead = DCB, ADE = CBD, de: BD = AE: CD, and de · DC = AE · BD