If E and F are the midpoints of equal sides AB and AC of a triangle ABC. Then:

BF=AC BF=AF CE=AB BF = CE

BF = CE

In In ΔEBC and ΔFCB, ∠EBC = ∠FCB (AB = AC) EB = FC (E and F are the mid points of equal sides) BC = BC (common side) ∴ ΔEBC ≅ ΔFCB (By SAS congruence condition) Thus, we can say BF = CE