∴△ABC≌△DBE

∴∠BCE=60，CE=BE，DE=AC

∫∠DCB = 30

∴∠DCE=90

In right-angle △DCE, according to Pythagoras theorem,

DC2+EC2=DE2，

∴DC2+BC2=AC2, that is, the quadrilateral ABCD is a Pythagorean quadrilateral. 4) Connect CE in (Figure 2), as shown in Figure 2, and rotate △ABC clockwise by 60 degrees around vertex B to get △DBE, which connects AD and DC, and ∠ DCB = 30.