If it is your method, you must prove that BCFE has four points.

But you can use the midline to solve the problem ~ first extend DA and EB to G, then the triangle OAB and ODE are regular triangles, and then extend DA and FC to H, so that OH=OD=2 can prove that G and H coincide. In triangles GED and GFD, B and C are the midpoint of GE and GF, so BC is the midline of triangles GE and GF, so BC//EF.