As shown in the figure, let EH‖BC pay AC to G, let C pay CH‖AB pay EH to H,

If f is SD‖BC, AB and CG are S and D respectively.

EBCH is a parallelogram with EH=BC.

ESFG and GFDH are isosceles trapezoid,

It can be proved that the triangle EGF and FDH are congruent.

So EF = FH2EF = EF+FH & gt；; =EH=BC？

So EF>= 1/2BC? (E and F are equal signs when they are the midpoint of AB and AC respectively)