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 :)