Triangles BFE and BDC are similar.
==> BF/BD = EF/CD
==> BF = ( EF/CD) x BD ---(1)
Triangles DFE and DBA are similar.
==> FD/BD = EF/AB
==> FD = (EF/AB) x BD ----(2)
Adding (1) and (2),
BF + FD = ( EF/CD) x BD + (EF/AB) x BD = EFxBD[(1/CD) + (1/AB)
As BF + FD = BD, above can be written as
BD = EFxBD[(1/CD) + (1/AB)
Dividing both sides by BD,
1 = EFx( 1/y + 1/x) = EFx [(x + y)/xy]
==> EF = xy / (x + y)