Note that Fa is equivalent to a ⋁ XFa, so that a<->Fa is equivalent to a<->(a ⋁ XFa). Now, if a is true, so is that equivalence, and if a is false, then we get 0<->(0 ⋁ XFa) which is !XFa, i.e., XG!a.

You can also argue like this:

a<->(a ⋁ XFa)
= a⋀(a ⋁ XFa) ⋁ !a⋀!(a ⋁ XFa)
= a ⋁ !a⋀!(a ⋁ XFa)
= a ⋁ !a⋀!a ⋀ !XFa
= a ⋁ !a⋀!XFa
= a ⋁ !a⋀XG!a
= a ⋁ XG!a