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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

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B

Parallelogram

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C

Kite

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D

Rectangle

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Solution

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 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.


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