The vertical line that passes through point C and is AB intersects with F, so ∠ 1+∠ 2 = 60 because △ABC is an isosceles triangle with three lines in one place.

∠ DCE = 60 = ∠ 3+∠ 2, so∠1=∠ 3.

The perpendicular line of CD passing through point E intersects CD at G, and the perpendicular line of AB passing through point D intersects AB at H, then

∠4+∠5=90，∠4+∠DCE=90，∠4=30

So ∠ 5 = ∠ DCE = 60, CD//AB.

△ABC, ∠ACB= 120 degrees, ∠ CAB = ∠ B = 30 degrees.

Because CD//AB, ∠ 3 = ∠ cab = 30.

∠ 1=∠3, so∠1= 30.

Due to DH//GE, ∠EDH=∠DEG.

∠ADH=30，？ ∠ADE=∠ADH+∠EDH

∠4=30 ,? ∠DEC=∠DEG+∠4

So? ∠ADE=？ ∠DEC, AD//CE can be introduced.

Because CD//AB, AD//CE, quadrilateral ACDE is a parallelogram.

In △ADH, DH= 1/2AD, DH= 1/2DE, so AD=DE.

In △BCE, CE= 1/2BE, and since CE=AD, BE=2AD=AD+DE.