In a ΔABC, the line joining the mid points D and E of sides AB and AC respectively, is produced to meet at point F such that DE = EF. What can you say about the figure DFCB?
Consider the triangles ΔADE and ΔCFE
DE = EF (given)
∠AED = ∠CEF (vertically opposite angles)
AE = CE (E is the mid point of AC)
Thus, ΔADE ≅ CFE (by SAS congruency)
∴ AD = CF (By CPCT)
But, AD = DB (D is the mid point of AB)
∴ DB = CF (1)
Also, ∠DAE = ∠FCE (By CPCT)
∴ AB || CF (Alternate angles are equal) (2)
Hence, from (1) and (2), we have,
DB || CF and DB = CF
Thus, DFCB is a parallelogram.