Question

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?

• A. Rhombus
• B. Parallelogram
• C. Kite
• D. Rectangle

Answer 1

The correct option is B

Parallelogram

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 concruency)

$\therefore$ AD = CF (by CPCT)

But, AD = DB (D is the mid point of AB)

$\therefore$ DB = CF (1)

Also, ∠DAE = ∠FCE (by CPCT)

$\therefore$ AB || CF (alternate angles are equal) (2)

Hence, from (1) and (2), we have,
DB || CF and DB = CF
Thus, DFCB is a parallelogram.