Given : E and F are the midpoints of AB and AC respectively.

CD II AB

in triangles AEF and CDF,

angle AFE = angle CFD (vertically opposite angles)

AF = CF (F is the midpoint of AC)

angle EAF = angle FCD (alternate interior angles)

therefore, triangles AEF and CDF are congruent by ASA congruence.

therefore, EF = DF (CPCT)

AE = EB = CD (CPCT)

In quadrilateral BCDE,

EB = CD and CD II BA

therefore, BCDE is a parallelogram (since a quadrilateral is a parallelogram if a pair of opposite sides are equal

and parallel)

therefore, EF II BC

hence, Proved.

hope it helped :)