It is known that in trapezoid ABCD, ab ∥ CD, ad = DC = BC = 4 ∠ DCB = 120 ° CE ⊥ AB, the area of AC ⊥ BC and trapezoid is calculated

It is known that in trapezoid ABCD, ab ∥ CD, ad = DC = BC = 4 ∠ DCB = 120 ° CE ⊥ AB, the area of AC ⊥ BC and trapezoid is calculated

In RT △ BCE, ∠ BCE = 90-60 = 30, so be = BC / 2 = 2, CE = √ 3bE = 2 √ 3aD = BC, trapezoid is isosceles trapezoid, ∠ DAB = ∠ B = 60ad = CD, so ∠ DAC = ∠ dcaab ‖ CD, so ∠ DCA = ∠ BAC, so ∠ DAC = ∠ BAC / 2 = 30, so ∠ ACB = 180 - ∠ B

In rectangular ABCD, the diagonal lines AC and BD intersect at point O, and the extension line of CE / / BD intersects AB through vertex C at e, and AC = CE. It is proved that the quadrilateral ABCD is a rectangle
Pictures.
Wrong number for sorry. Change the first rectangle to a quadrilateral

Because AE ∥ DC, CE ∥ BD,
So the quadrilateral cebd is a parallelogram,
So be = CD = AB,
And because AC = CE,
So BC ⊥ AE,
So the quadrilateral ABCD is parallelogram