The isosceles triangle ABC, with a point D on the waist AB, connects DC and makes the isosceles triangle EDC with DC as the bottom edge, which is similar to the triangle ABC, connects AE and proves that AE is parallel to BC

The isosceles triangle ABC, with a point D on the waist AB, connects DC and makes the isosceles triangle EDC with DC as the bottom edge, which is similar to the triangle ABC, connects AE and proves that AE is parallel to BC

∵△EDC∽△ABC
∴∠DCE=∠BCA
However, ACE = DCE - ACD - BCD = BCA - ACD
Get ∠ ace = ∠ BCD
Also ∵ △ EDC ∽ ABC
∴EC:AC＝DC:BC
After a transformation, EC: DC = AC: BC plus ∠ ace = ∠ BCD
We can get △ ace ∽ BCD
∴∠CAE=∠B
∵AB=AC
∴∠CAE=∠B=ACB
∴∠CAE+∠BAC+∠B=∠ACB+∠BAC+∠B=180°
That is, EAB + B = 180 degree
∴AE∥BC