This is from chapter Quadrilaterals. Plz wrie everything given, To prove.
ABCD is a parallelogram.
BE = AB
ED bisects BC
AB = BE (Given)
AB = CD (Opposite sides of ||gm)
∴ BE = CD
Let DE intersect BC at F.
In ΔCDO and ΔBEO,
∠DCO = ∠EBO (AE || CD)
∠DOC = ∠EOB (Vertically opposite angles)
BE = CD (Proved)
ΔCDO ≅ ΔBEO by AAS congruence condition.
Thus, BF = FC (by CPCT)
Therefore, ED bisects BC. Proved