E is the mid-point of side AD of a parallelogram ABCD. A line through A parallel to EC meets BC at F and DC produced at G. Prove that
(i) DG = 2AB
(ii) AG = 2AF

In ΔADG, E is the mid-point of AD and EC ïï AG
.^.. EC = AG and EC ïï AG ⇒  2EC = AG ...(i)
and C is the mid-point of DG.
.^.. DG = DC + CG = 2DC...(ii)
but,  DC = AB ...(opposite sides of a parallelogram)
.^.. DG = 2AB  
(ii) In AFCE, AE ïï FC and AF ïï EC,
.^.. Quadrilateral AFCE is a parallelogram 
.^.. EC = AF...(iii)
From equations (i) and (iii) we get,
2AF = AG ⇒ AG = 2AF