If we shift a body in equilibrium from A to C in a gravitational field via path AC or ABC : 

The work done bt the force \(\vec{F}\) for both paths will be same

For the path AC :

W_AC = Fs cos (90^o - θ)

  = mgs sin θ 

  = mgh

For the path AB :

W_AB = Fa cos 90^o 

   = 0

For the path BC :

W_BC = Fh cos 0^o 

  = mgh

So that : W_AB + W_BC = mgh = W_AC

i.e. W_ABC = W_BC