Given ABCD is a parallelogram, BC is produced to F and BD is produced to E and AE=CF Prove that:
(i) BE || DF
(ii) BD and EF bisect each other. 

To Prove: BE ïï DF
and EO = OF
and DO = OBProof: In ΔABE and ΔCDF
    AB = DC ...... (opposite sides of parallelogram)
     AE = CF ...... (given)
and BE = DF ...... (given)
... ΔABE ≅ ΔCDF
... ∠ABE = ∠CDF ...... (i) (c.p.c.t.)
Now since AB ïï DC and DB is a transversal
... ∠ABD = ∠BDC ...... (ii) (Alternate interior angles)
Adding equations (i) and (ii), we get
∠ABE + ∠ABD = ∠CDF + ∠BDC
       ⇒ ∠EBD = ∠BDF ⇒ BE ïï DF
Now in ΔOBE and ΔODF,  
              BE = DF ...... (given)
         ∠EBO = ∠FDO ...... (proved above)
    and ∠BOE = ∠DOF ...... (Vertically opposite angles)
        .^..ΔOBE ≅ ΔODF
          .^.. OE = OF
        and OB = OD ...... (c.p.c.t.)
Therefore O is the mid-point of EF and DB.